Mathematics for the international student Mathematics HL (option): Calculus

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Author: Catherine Quinn | Chris Sangwin | Robert Haese | Michael Haese

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HAESE MATHEMATICS Specialists in mathematics publishing

Mathematics

for the international student

Mathematics HL (Option): Calculus

HL Topic 9 FM Topic 5 Catherine Quinn Chris Sangwin Robert Haese Michael Haese

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for use with IB Diploma Programme black

Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_00\001IB_HL_OPT-Calculus_00.cdr Tuesday, 19 March 2013 2:00:41 PM BRIAN

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MATHEMATICS FOR THE INTERNATIONAL STUDENT Mathematics HL (Option): Calculus Catherine Quinn Chris Sangwin Robert Haese Michael Haese

B.Sc.(Hons), Grad.Dip.Ed., Ph.D. M.A., M.Sc., Ph.D. B.Sc. B.Sc.(Hons.),Ph.D.

Haese Mathematics 152 Richmond Road, Marleston, SA 5033, AUSTRALIA Telephone: +61 8 8210 4666, Fax: +61 8 8354 1238 Email: [email protected] www.haesemathematics.com.au Web: National Library of Australia Card Number & ISBN 978-1-921972-33-1 © Haese & Harris Publications 2013 Published by Haese Mathematics. 152 Richmond Road, Marleston, SA 5033, AUSTRALIA First Edition

2013

Artwork by Brian Houston and Gregory Olesinski. Cover design by Piotr Poturaj. Typeset in Australia by Deanne Gallasch. Typeset in Times Roman 10 \Qw_ . Printed in Hong Kong through Prolong Press Limited. The textbook and its accompanying CD have been developed independently of the International Baccalaureate Organization (IBO). The textbook and CD are in no way connected with, or endorsed by, the IBO. This book is copyright. Except as permitted by the Copyright Act (any fair dealing for the purposes of private study, research, criticism or review), no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publisher. Enquiries to be made to Haese Mathematics. Copying for educational purposes: Where copies of part or the whole of the book are made under Part VB of the Copyright Act, the law requires that the educational institution or the body that administers it has given a remuneration notice to Copyright Agency Limited (CAL). For information, contact the Copyright Agency Limited. Acknowledgements: While every attempt has been made to trace and acknowledge copyright, the authors and publishers apologise for any accidental infringement where copyright has proved untraceable. They would be pleased to come to a suitable agreement with the rightful owner.

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Disclaimer: All the internet addresses (URLs) given in this book were valid at the time of printing. While the authors and publisher regret any inconvenience that changes of address may cause readers, no responsibility for any such changes can be accepted by either the authors or the publisher.

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FOREWORD Mathematics HL (Option): Calculus has been written as a companion book to the Mathematics HL (Core) textbook. Together, they aim to provide students and teachers with appropriate coverage of the two-year Mathematics HL Course, to be first examined in 2014. This book covers all sub-topics set out in Mathematics HL Option Topic 9 and Further Mathematics HL Topic 5, Calculus. The aim of this topic is to introduce students to the basic concepts and techniques of differential and integral calculus and their applications. Detailed explanations and key facts are highlighted throughout the text. Each sub-topic contains numerous Worked Examples, highlighting each step necessary to reach the answer for that example. Theory of Knowledge is a core requirement in the International Baccalaureate Diploma Programme, whereby students are encouraged to think critically and challenge the assumptions of knowledge. Discussion topics for Theory of Knowledge have been included on pages 129 and 140. These aim to help students discover and express their views on knowledge issues. The accompanying student CD includes a PDF of the full text and access to specially designed graphing software. Graphics calculator instructions for Casio fx-9860G Plus, Casio fx-CG20, TI-84 Plus and TI-nspire are available from icons located throughout the book. Fully worked solutions are provided at the back of the text, however students are encouraged to attempt each question before referring to the solution. It is not our intention to define the course. Teachers are encouraged to use other resources. We have developed this book independently of the International Baccalaureate Organization (IBO) in consultation with experienced teachers of IB Mathematics. The Text is not endorsed by the IBO. In this changing world of mathematics education, we believe that the contextual approach shown in this book, with associated use of technology, will enhance the students understanding, knowledge and appreciation of mathematics and its universal applications. We welcome your feedback. Email:

[email protected]

Web:

www.haesemathematics.com.au

CTQ

CS

RCH PMH

ACKNOWLEDGEMENTS

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The authors and publishers would like to thank all those teachers who offered advice and encouragement on this book, with particular mention to Peter Blythe.

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USING THE INTERACTIVE STUDENT CD The interactive CD is ideal for independent study. Students can revisit concepts taught in class and undertake their own revision and practice. The CD also has the text of the book, allowing students to leave the textbook at school and keep the CD at home. By clicking on the relevant icon, a range of interactive features can be accessed: w Graphics calculator instructions for the Casio fx-9860G Plus, Casio fxCG20, TI-84 Plus and the TI-nspire w Interactive links to graphing software

INTERACTIVE LINK

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GRAPHICS CALCUL ATOR INSTRUCTIONS

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TABLE OF CONTENTS

5

TABLE OF CONTENTS

SYMBOLS AND NOTATION USED IN THIS BOOK A B C D E F G H

6

Number properties Limits Continuity of functions Differentiable functions l’Hôpital’s Rule Rolle’s theorem and the Mean Value Theorem (MVT) Riemann sums The Fundamental Theorem of Calculus Z 1 f (x) dx Improper integrals of the form

I

9 12 20 25 29 34 38 44 50

a

Sequences Infinite series Taylor and Maclaurin series Differential equations Separable differential equations The integrating factor method Taylor or Maclaurin series developed from a differential equation Review set A Review set B Review set C Review set D

57 66 89 105 112 117 120 123 124 125 126

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INDEX

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WORKED SOLUTIONS

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141

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APPENDIX B (Formal definition of a limit)

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140

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THEORY OF KNOWLEDGE (Axioms and Occam’s razor)

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APPENDIX A (Methods of proof)

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THEORY OF KNOWLEDGE (Torricelli’s trumpet)

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J K L M N O P

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6

SYMBOLS AND NOTATION USED IN THIS BOOK

¼ > > < 6

is is is is is

f......g 2 2 = N Z Q R Z+ µ ½ ) ) Á f : A!B f : x 7! y

approximately equal to greater than greater than or equal to less than less than or equal to

the set of all elements ...... is an element of is not an element of the set of all natural numbers the set of integers the set of rational numbers the set of real numbers the set of positive integers is a subset of is a proper subset of implies that does not imply that f is a function under which each element of set A has an image in set B f is a function under which x is mapped to y

f (x)

the image of x under the function f

f ±g

or f (g(x)) the composite function of f and g

jxj

the modulus or absolute value of x

[a, b]

the closed interval a 6 x 6 b

] a, b [

the open interval a < x < b the nth term of a sequence or series

un fun g

the sequence with nth term un

Sn S1 n P ui

the sum of the first n terms of a sequence the sum to infinity of a series u1 + u2 + u3 + :::: + un

i=1 n Q

u1 £ u2 £ u3 £ :::: £ un

ui

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IB HL OPT 2ed Calculus


7

lim f(x)

the limit of f(x) as x tends to a

lim f (x)

the limit of f(x) as x tends to a from the positive side of a

lim f (x)

the limit of f(x) as x tends to a from the negative side of a

maxfa, bg 1 P cn xn

the maximum value of a or b

x!a

x!a+

x!a¡

the power series whose terms have form cn xn

n=0

dy dx f 0 (x)

the derivative of y with respect to x the derivative of f (x) with respect to x

2

d y dx2 f 00 (x) dn y dxn f (n) (x) R y dx Rb y dx a

the second derivative of y with respect to x the second derivative of f (x) with respect to x the nth derivative of y with respect to x the nth derivative of f (x) with respect to x the indefinite integral of y with respect to x

the definite integral of y with respect to x between the limits x = a and x = b ex exponential function of x ln x the natural logarithm of x sin, cos, tan the circular functions csc, sec, cot the reciprocal circular functions arcsin, arccos, arctan the inverse circular functions A(x, y) the point A in the plane with Cartesian coordinates x and y [AB]

the line segment with end points A and B

AB

the length of [AB]

b A

the angle at A

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is parallel to

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is perpendicular to

5

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the angle between [CA] and [AB] the triangle whose vertices are A, B, and C

100

50

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b or ]CAB CAB 4ABC k ?

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9

CALCULUS

A

NUMBER PROPERTIES

IMPORTANT NUMBER SETS You should be familiar with the following important number sets: ² Z + = f1, 2, 3, .... g is the set of positive integers. ² N = f0, 1, 2, 3, .... g is the set of natural numbers. ² Z = f...., ¡2, ¡1, 0, 1, 2, .... g is the set of integers. p

² Q is the set of rational numbers. These are numbers which can be expressed in the form where q p, q 2 Z , q 6= 0. ² R is the set of real numbers comprising the rational numbers Q , and the irrational numbers which lie on the number line but cannot be expressed as a ratio of integers. The number sets follow the hierarchy Z + ½ N ½ Z ½ Q ½ R . In this option topic we will be principally concerned with the set R . Rigorous treatments of the algebraic and set theoretic properties of R , such as the fact that R is a continuous set, are available in a variety of calculus and analysis books. However, we will outline here only those results of most immediate relevance to our work with limits, sequences, and series. A closed interval consisting of all real numbers from a to b inclusive is denoted [a, b]. [a, b] is fx j a 6 x 6 bg. An open interval consisting of all real numbers between a and b is denoted ] a, b [ . ] a, b [ is fx j a < x < bg.

a

b

a

b

THE ABSOLUTE VALUE FUNCTION y

For any a 2 R , the absolute value of a, denoted by jaj, is defined by: ½ a if a > 0 jaj = ¡a if a < 0

y = |x| x

The absolute value jaj of a number a 2 R is the distance from a to the origin on the real number line. More generally, the distance between two numbers a, b 2 R on the number line is given by ja ¡ bj. The absolute value function has the following properties: 1 jaj > 0 for all a 2 R . 2 j¡aj = jaj for all a 2 R . 3 jabj = jaj jbj for all a, b 2 R . 4 a = jaj or ¡jaj, and hence ¡jaj 6 a 6 jaj for all a 2 R .

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5 If c > 0 then jaj 6 c if and only if ¡c 6 a 6 c.

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10

CALCULUS

We will also need the property of real numbers that: If a 6 c and b 6 d, then a + b 6 c + d, for a, b, c, d 2 R . Proof of Property 5: ) Suppose that jaj 6 c. Since a 6 jaj and ¡a 6 jaj, we find a 6 jaj 6 c and ¡a 6 jaj 6 c ) a 6 c and ¡a 6 c.

For a proof “if and only if ” we prove it one way ) and then the other (.

But ¡a 6 c is equivalent to ¡c 6 a, so combining the two inequalities we have ¡c 6 a 6 c. ( If ¡c 6 a 6 c, then a 6 c and ¡c 6 a: ) ¡a 6 c: ) since jaj = a or ¡a, jaj 6 c.

THE TRIANGLE INEQUALITY The Triangle Inequality states: For any a, b 2 R , ja + bj 6 jaj + jbj. Proof: From Property 4 we have ¡jaj 6 a 6 jaj and ¡jbj 6 b 6 jbj for all a, b 2 R . Adding these inequalities gives ¡(jaj + jbj) 6 a + b 6 jaj + jbj. Using Property 5 with c = jaj + jbj, this is equivalent to ja + bj 6 jaj + jbj. Corollaries: 1 ja ¡ bj 6 jaj + jbj for all a, b 2 R . 2 jaj ¡ jbj 6 ja + bj for all a, b 2 R . 3 jaj ¡ jbj 6 ja ¡ bj for all a, b 2 R . Proofs: 1 By the Triangle Inequality, we have ja + cj 6 jaj + jcj for all a, c 2 R . ) letting c = ¡b, we get ja ¡ bj 6 jaj + j¡bj for all a, b 2 R . ) ja ¡ bj 6 jaj + jbj jaj = j(a + b) + (¡b)j

2 )

jaj 6 ja + bj + j¡bj )

jaj 6 ja ¡ bj + jbj

for all a, b 2 R

by the Triangle Inequality.

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jaj ¡ jbj 6 ja ¡ bj

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by the Triangle Inequality.

jaj = j(a ¡ b) + bj

3

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jaj ¡ jbj 6 ja + bj

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CALCULUS

EXERCISE A 1 Prove that jaj > 0 for all a 2 R . 2 Prove that j¡aj = jaj for all a 2 R . 3 Prove that ja1 + a2 + .... + an j 6 ja1 j + ja2 j + .... + jan j for any a1 , a2 , ...., an 2 R . 4 If a < x < b and a < y < b show that jx ¡ yj < b ¡ a. Interpret this result geometrically. 5 Prove that ja ¡ bj 6 ja ¡ cj + jc ¡ bj. 6 Prove that if jx ¡ aj
.

7 If jx ¡ aj < " and jy ¡ bj < " show that j(x + y) ¡ (a + b)j < 2". 8 The Archimedean Property states that for each pair of positive real numbers a and b, there is a natural number n such that na > b. Use the Archimedean Property to prove that for each positive number " there is a natural number n such that

1 < ". n

The properties of R in questions 8 and 9 are needed later in the course.

9 Prove the Bernoulli Inequality by mathematical induction: If x > ¡1 then (1 + x)n > 1 + nx for all n 2 Z + .

10 The Well-Ordering Principle states that every non-empty subset of Z + has a least element. Show that the Well-Ordering Principle does not apply to R + , the set of positive real numbers. 11 If r 6= 0 is rational and x is irrational, prove that r + x and rx are irrational.

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For questions 10 and 11 you will need to write a proof by contradiction. For help with this, consult Appendix A: Methods of proof.

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11

12

CALCULUS

B

LIMITS

Consider a real function f(x) whose domain D is a subset of R . We can write f : D ! R .

1 is not a real number. x ! 1 refers to positive values of x becoming increasingly large.

We wish to examine the behaviour of the function:

² as x approaches some particular finite value a 2 R , so x ! a ² as x tends to infinity or negative infinity, so x ! 1 or x ! ¡1, when D = R .

We shall work with the following informal definition of a limit of a function. For more information, consult Appendix B: Formal definition of a limit. Let a 2 R be a fixed real number. Let f be a function defined in an open interval about x = a, except f (a) need not be defined. We say the number l is the limit of f as x approaches a, provided f(x) becomes as close as we like to l by choosing values of x close enough, but not equal to, the number a. We write lim f(x) = l. x!a

From this definition, we see that f (x) gets closer and closer to l as x gets closer and closer to a, from either side of a. If the function does not approach a finite value l, we say the limit does not exist (DNE). If f (x) gets closer and closer to l as x gets closer and closer to a from the right of a (where x > a), we say “x approaches a from the right”, and write lim f(x) = l. x!a+

If f (x) gets closer and closer to l as x gets closer and closer to a from the left of a (where x < a), we say “x approaches a from the left”, and write lim f (x) = l. x!a¡

For a 2 R , lim f (x) exists if and only if both

lim f(x) and

x!a

x!a¡

In this case lim f (x) = lim f (x) = lim f (x). x!a

x!a¡

lim f (x) exist and are equal.

x!a+

x!a+

For example: lim (x + 1) = 2

y

x!1

y=x+1

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In this case f(1) can be evaluated directly, but we do not do this. Instead, when calculating limits we consider the behaviour of f(x) for x close to 1.

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CALCULUS

x2 ¡ 1 x¡1

² y =

is not defined at x = 1. This is not a

y

problem, since in determining the limit as x approaches 1, we never let x actually reach the value 1. But as we make x very, very close to value 1, we can make f (x) as close as we like to value 2. x2 ¡ 1 (x ¡ 1)(x + 1) = lim x!1 x ¡ 1 x!1 (x ¡ 1)

y=

2

x2 ¡ 1 x¡1

1

fsince x 6= 1g

lim

13

1

x

= lim x + 1 x!1

=2 ²

y y=

As x ! 1¡ , f(x) ! ¡1. As x ! 1+ , f(x) ! 1.

1 x¡1

There is no finite real number that f (x) approaches as x approaches value 1, so the limit of f as x approaches 1 does not exist (DNE).

x

-1

½ ² Consider f(x) =

x!1

1 x¡1

DNE.

x + 1, x 6 1 x > 1. x2 ,

As x ! 1¡ , f(x) ! 2 and so

y

lim f(x) = 2.

y = f(x)

x!1¡

As x ! 1+ , f(x) ! 1 and so Since

lim

So,

x=1

lim f (x) = 1.

x!1+

2 1

lim f (x) 6= lim f(x), there is no unique

x!1¡

x!1+

value which the function approaches as x ! 1. ) lim f (x) DNE.

1

x

x!1

We can make f(x) as close as we like to value 5 by making x large enough. As x ! 1, f(x) ! 5 from below. We write lim f(x) = 5¡ .

y

5

y =5-

x!¡1

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Similarly, as x ! ¡1, f (x) ! 5 from above. We write lim f(x) = 5+ .

1 x

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14

CALCULUS

Example 1 By examining the graphs of y = x, y = a

lim x DNE

a

lim

b

x!1

x!1

1 , x

1 , establish the following limits: x2 1 1 c lim 2 = 0 d lim DNE x!1 x x!0 x

and y =

1 =0 x

b

y

y y=

y=x

1 x

x x

As x ! 1, y = x ! 1 ) lim x DNE. x!1

) y

c

lim

x!1

1 = 0. x

d y=

1 ! 0+ x

As x ! 1,

y

1 x2

y=

1 x

x

x

As x ! 1, )

1 ! 0+ x2

As x ! 0+ ,

1 = 0. x!1 x2

As x ! 0¡ ,

lim

)

1 x

lim

x!0

1 !1 x 1 ! ¡1 x

DNE.

EXERCISE B.1 1

1

1 By examining the graphs of y = ¡x, y = ¡ , and y = ¡ 2 , establish the following limits: x x ³ ´ ³ ´ ³ ´ 1 1 1 a lim (¡x) DNE b lim ¡ =0 c lim ¡ 2 = 0 d lim ¡ DNE x!1

x!1

x!1

x

x!0

x

x

2 Sketch each function and determine the existence or otherwise of lim f (x). x!a

x2 + x ¡ 2 b f (x) = , x+2

a f (x) = 3x + 2, a = ¡1

( a Sketch the function f (x) =

lim f (x) and ¼¡

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2x , ¼

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sin x, x >

a = ¡2

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CALCULUS

15

c Determine the existence or otherwise of lim f (x): ¼

x! 2

a Sketch the function f(x) =

4

x+1 . x¡1

b Determine the existence or otherwise of: i lim f(x) ii lim f (x)

iii

x!1

x!¡1

8 2 < x , x>3 5 Let f (x) = 5, x=3 : 3x, x < 3. a Sketch the graph of y = f(x).

b Evaluate

lim f (x)

x!1

lim f (x) and

x!3¡

lim f (x).

x!3+

c Does lim f(x) exist? Explain your answer. x!3

p 6 Does lim x exist? Explain your answer. x!0

7 For each of the following functions, discuss the limits: lim f (x)

i

½ a f(x) =

lim f(x)

ii

x!0¡

3

(x ¡ 1) , x 6 0 1, x>0

c f(x) = sin

iii

x!0+

b f(x) =

³ ´ 1 , x 6= 0

8 < sin x,

lim f (x)

x!0

x0

x

THE LIMIT LAWS We often define new functions using a sum, composition, or some other combination of simpler functions. The limit laws help us calculate limits for these new functions using what we already know about the simpler functions. For information about proving these laws, consult Appendix B: Formal definition of a limit. If f (x) = c a constant, where c 2 R , then

lim f (x) = lim c = c, for all a 2 R .

x!a

x!a

We can state the limit laws as follows: Consider real functions f(x) and g(x) for which lim f (x) = l and lim g(x) = m, x!a x!a where a, l, m 2 R .

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g(x)

m

for all n 2 Z + lim f (x) = l p p lim n f (x) = n l for all n 2 Z + provided l > 0 n

n

x!a

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²

lim f (x) g(x) = lm µ ¶ f (x) l lim = provided m 6= 0

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for any constant c 2 R

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²

lim cf (x) = cl

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16

CALCULUS

From the first limit law we see that multiplying by the constant 1 will not change the value or existence of a limit. When evaluating many limits, it often helps to multiply by 1 in a well-chosen form. We will use this technique for rational functions.

RATIONAL FUNCTIONS f(x) is a rational function if it can be written in the form f(x) =

p(x) , where p, q are real polynomials. q(x)

Suppose xm is the highest power of x present in either p or q. To examine lim

x!1

p(x) p(x) or lim , we divide all terms in the numerator and denominator by xm . q(x) x!¡1 q(x) 1

m p(x) p(x) = lim £ x1 x!1 q(x) x!1 q(x)

lim

1 xm 1 xm

since

xm

= 1 for all x 6= 0 and multiplying by the constant 1

does not change the value of the limit. For rational functions, this allows us to make use of the known limits

lim

x!§1

1 = 0 for all m 2 Z + . xm

Example 2 Determine the existence or otherwise of the following limits: a

x+5 x!1 ¡2x2 + x + 1

lim

lim

a

x!1

b

x+5 ¡2x2 + x + 1

10x2 ¡ 5 x!1 3x2 + x + 2

lim

c

lim

b

x!1

1

x!1

=

0

+

¡2 +

+

= lim

1 x2

x!1

fas x ! 1,

¡2

x2

5 x2

1 x

1 x

1

= lim

x2

= lim

10x2 ¡ 5 3x2 + x + 2

2 10x2 ¡ 5 £ x1 2 x!1 3x + x + 2

2 x+5 £ x1 2 x!1 ¡2x + x + 1

= lim

1 x

x2 + x + 1 x!1 x¡2

lim

! 0 and

1 x2

=

! 0g

10 3

10 ¡

5 x2

1 3+ x + x22

=0 c

1 1 1+ x + x12 2 x2 + x + 1 x2 + x + 1 = lim £ x1 = lim 1 x!1 x¡2 x!1 x¡2 x!1 ¡ x22 x x2

lim

As x ! 1, the numerator ! 1, but the denominator ! 0. Hence as x ! 1, )

x2 + x + 1 x!1 x¡2

x2 + x + 1 ! 1. x¡2

lim

DNE.

EXERCISE B.2 1 Evaluate the following limits, where possible:

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1 1 ¡ y 5

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µ

x2 + 3x ¡ 4 x!0 x¡1

c lim

1 f lim y !5 y ¡ 5

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x+1 ¡ 2x ¡ 3

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x!¡1 x2

1 e lim y !2 y ¡ 5

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x2 + 3x ¡ 4 d lim x!1 x¡1

lim

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x+1 ¡ 2x ¡ 3

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x!1 x2

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a lim

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CALCULUS

17

2 Evaluate the following limits, where possible: a

ln x 2¡x

lim p

x!2¡

d lim

µ!0

lim

b

cos µ µ

x!0

sin x ex

¼¡ x! 2

3 Evaluate the following limits, where possible:

r

4x2 ¡ 5x + 1 b lim x!1 x2 + x + 1

3+x a lim x!1 x + 5

lim

c

x!1

p

x2 + x + x

p

4 By first multiplying by

x2

+x+x

lim

x!¼ ¡

sin x 1 ¡ cos x

tan x sec x

lim

e

c

lim

, find

x!1

x2 + 3 x

d

lim

x!1 x2

1+x +x+1

p x2 + x ¡ x. p

x2 + x + x

p

x2 + x + x

5 Let f be a function with domain R , and let a, l 2 R lim f(x) = l

Use the limit laws to prove that

be constants. Suppose lim f (x) exists. x!a

if and only if

x!a

= 1, for x 6= 0.

lim (f(x) ¡ l) = 0.

x!a

THE SQUEEZE THEOREM The Squeeze Theorem shows us that inequalities between functions are preserved when we take limits. Let f, g, h be real functions and let a, l 2 R . Suppose that f (x) 6 g(x) 6 h(x) for all x 6= a in some open interval containing a. If

lim f (x) = l = lim h(x) then

x!a

lim g(x) = l .

x!a

x!a

So, a function g(x) is forced to have the same limit as f and h if g is squeezed between them. Example 3 Use the Squeeze Theorem to evaluate

lim x2 cos

x!0

³ ´ 9 . x

³ ´ 9 x

6 1 for all x 2 R , ³ ´ 9 6 x2 ¡x2 6 x2 cos

Since ¡1 6 cos

x

2

lim ¡x = 0 = lim x2 .

Now

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x!0

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18

CALCULUS

Consider the limit lim

x!0

sin x . Since lim sin x = 0 and lim x = 0, we say this limit has indeterminate x!0 x!0 x

form 00 . We cannot use the limit laws to determine whether or not this limit exists, or what its value might be. However, one way to evaluate the limit is to use the Squeeze Theorem. lim

The Fundamental Trigonometric Limit is

Proof:

Consider the unit circle with angle 0 < µ < ¼2 , and also points P(cos µ, sin µ), Q(cos µ, 0), R(1, 0), and S(1, tan µ).

y 1

S (1, tan µ)

P

µ! 0

sin µ = 1. µ

µ Q R1

x

Clearly, area of 4OPQ < area of sector OPR < area of 4ORS 1 2

)

cos µ sin µ < 12 µ < 1 2

Since 0 < µ < ¼2 ,

1 2

sin µ cos µ

sin µ > 0

µ 1 < sin µ cos µ 1 sin µ > > cos µ ) cos µ µ sin µ 1 < ) cos µ < µ cos µ

) cos µ
0 ¡1, x < 0 y

1 x¡5

y = f(x)

1

x

x -1 x=5

f is not defined at x = 5, and lim f(x) DNE.

f is defined for all x, but there is a ‘jump’ discontinuity at x = 0.

x!5

lim f (x) = ¡1 and

f is continuous for all x 2 R , x 6= 5. f has an essential discontinuity at x = 5.

x!0¡

)

lim f (x) = 1

x!0+

lim f(x) DNE.

x!0

f is continuous for all x 2 R , x 6= 0. f has an essential discontinuity at x = 0. Example 5 Discuss the continuity of the following functions. If there is a removable discontinuity, describe how this could be removed. ( sin x ½ 2 , x 6= 0 x , x 6= 2 x a f (x) = b f (x) = 6, x = 2 0, x=0 a

f is defined for all x, but there is a discontinuity at x=2 lim f (x) = 4 and lim f(x) = 4

y 6

y = f(x)

x!2¡

4

x!2+

lim f(x) = 4

) But

lim f(x) 6= f (2), so there is a removable

x!2

discontinuity when x = 2.

x

2

x!2

f is continuous for all x 2 R , x 6= 2.

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The discontinuity can be removed by defining a new function based on f, but which is continuous at x = 2. ½ 2 x , x= 6 2 This is g(x) = which is actually just g(x) = x2 . 4, x=2

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22

CALCULUS

b

f is defined for all x, but there is a discontinuity at x = 0. lim f (x) = 1 and lim f (x) = 1

y 1 y = f(x)

x!0¡

x!0+

lim f (x) = 1

) x

But

x!0

lim f(x) 6= f (0), so there is a removable

x!0

discontinuity when x = 0. f is continuous for all x 2 R , x 6= 0. The discontinuity can be removed by defining a new function based on f , but which is ( sin x , x= 6 0 x continuous at x = 0. This is g(x) = 1, x = 0.

EXERCISE C.1 1 Suppose f and g are functions which are continuous at x = a. Use the limit laws to prove that the following functions are also continuous at x = a: a f (x) g(x)

b f (x) § g(x)

f (x) , g(x)

c

for g(a) 6= 0

e [f (x)]n , n 2 Z +

d c f (x), for c 2 R a constant 2 Consider the function f with graph:

y 6

b For which values of 8 x, > > > > x2 ¡ 6, > > > > < 5, 3 Let f (x) = 6, > > > > 5 , > > > 4¡x > : 2x,

y = f(x)

4

a Complete the following: i f has an essential discontinuity at x = :::::: ii f has a removable discontinuity at x = ::::::

1

2

x

5 x=6

x 2 R is f continuous? x < ¡2 ¡2 6 x < 0 06x 3 x ¡ 10x + 7, x > 3 a f (x) = 5, b g(x) = x=3 : 5, x 0 k + 1, x < 2 x c f(x) = d f(x) = : k, x 6 0 x2 , x>2 ½ ½ kx, x > 0 k + x, x>2 e f(x) = f f(x) = 0, x60 jk + 2j , x < 2 6 For a real function f with domain D, an alternative (equivalent) definition of continuity is as follows: “f is continuous at a 2 D if, for any sequence fxn g of values in D such that lim xn = a, then lim f(xn ) = f(a).” n!1 n!1 ½ 1, x 2 Q The Dirichlet function is defined by g(x) = 0, x 2 =Q. a Consider the sequence xn = a +

p 2 , a 2 Q , n 2 Z + . Use the given definition of continuity n

to show that g(x) is discontinuous at x = a. b Now suppose a 2 = Q , and that xn is the decimal expansion of a to n decimal places, n 2 Z + . Use the given definition of continuity to show that g(x) is discontinuous at x = a. c Hence discuss the continuity of the Dirichlet function. 7 Consider two functions f, g and two values a, l 2 R . Suppose f is continuous on R . If l = g(a), you may assume the result: “If g is continuous at a, and f is continuous at g(a), then f ± g is continuous at a.” lim f (g(x)) = f(l). ´ b Show that lim f (g(x)) = f lim g(x) whenever lim g(x) exists. x!a x!a x!a ³ ´ c Show that lim f(g(x)) = f lim g(x) whenever lim g(x) exists. a If

lim g(x) = l, show that

x!a

³

x!1

x!1

1

a Show that x n = eln x

8

x!a

1 n

x!1

for x > 0, x 2 R .

b Use the fact that f(x) = ex

is continuous on R to prove that

x 2 R , n 2 Z +.

1

lim x n = 1 for x > 0,

n!1

THE INTERMEDIATE VALUE THEOREM (IVT) A function f is continuous on a closed interval [a, b], a < b, if f is continuous at x for all x 2 ] a, b [ , and also lim f (x) = f(a) and lim f(x) = f(b). x!a+

x!b¡

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The following theorem formalises the intuitive property that a function f which is continuous on a closed interval [a, b] has no ‘breaks’ or ‘holes’ between x = a and x = b, and will in fact take every value between f(a) and f (b) as we increase x from a to b.

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24

CALCULUS

The Intermediate Value Theorem (IVT) states that: Suppose a function f is continuous on a closed interval [a, b]. If k is any value between f (a) and f (b), then there exists c 2 [a, b] such that f (c) = k. y

y

f(a)

f(b) k

k

or f(a)

f(b)

c

a

x

b

a

c

b

x

BOUNDED FUNCTIONS A function f is bounded on [a, b] if, for all x 2 [a, b], jf(x)j 6 M for some M 2 R . In other words, a function is bounded on [a, b] if it does not tend to infinity or negative infinity on the interval [a, b]. If a function f is continuous on [a, b], then f is bounded on [a, b]. It follows that if f is continuous on [a, b], a < b, then: ² f has a maximum value M on the interval [a, b] where M = f(xM ) for some xM 2 [a, b] ² f has a minimum value m on the interval [a, b] where m = f (xm ) for some xm 2 [a, b].

EXERCISE C.2 1 Using the IVT, explain why: a f (x) = x3 + x ¡ 3 has a real zero in the interval [0, 2] b f (x) =

4 x¡2

does not have a real zero in the interval [1, 3] even though f(1) < 0 and

f (3) > 0.

2 Consider the function f(x) = x3 ¡ 9x2 + 24x ¡ 10 on the interval [1, 5]. Find: a the maximum value M of f on [1, 5], and the values xM 2 [1, 5] such that f (xM ) = M b the minimum value m of f on [1, 5], and the values xm 2 [1, 5] such that f(xm ) = m. 3 Suppose f is continuous on [a, b], a < b, and suppose f(a), f (b) have opposite signs. Prove that f has at least one zero between a and b. a Prove that for each constant r 2 R , r > 0, there exists a real value of x such that sin x = 1 ¡ rx. Hint: Let f (x) = sin x + rx ¡ 1 and apply the IVT to a suitable interval.

4

b Suppose r = 1. Use your calculator to solve sin x = 1 ¡ x on the domain x 2 ] ¡2, 2 [ . a Use the IVT to prove that f (x) = x15 +

5

1 1 + sin2 x + (r + 1)2

has at least one zero on the

interval ] ¡1, 1 [ for any constant r 2 R .

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Explain your answer.

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] ¡1, 1 [ of g(x) = rx17 +

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b Let r 2 R , r > 0 be any constant. Can the IVT be used to prove the existence of a zero in

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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\024IB_HL_OPT-Calculus_01.cdr Friday, 15 March 2013 3:12:00 PM BRIAN

CALCULUS

D

25

DIFFERENTIABLE FUNCTIONS

Suppose f is a real function with domain D containing an open interval about x = a. f (a + h) ¡ f (a) exists. If this limit exists we denote it by h f (a + h) ¡ f (a) f 0 (a) = lim is the derivative of f at x = a. h h!0

f is differentiable at x = a if f 0 (a), and

lim

h!0

If f is differentiable at a for all a 2 D, then we say f is a differentiable function. By letting x = a + h in the definition of the derivative, and noting that as h ! 0, x = a + h ! a, we obtain the following alternative form for the derivative of f at x = a: f 0 (x) = lim

x!a

f (x) ¡ f (a) . x¡a

Example 6 lim

a Prove that

h!0

cos h ¡ 1 = 0. h

b Using the limit definition of the derivative, prove that if f(x) = sin x then f 0 (x) = cos x. cos h ¡ 1 cos h ¡ 1 cos h + 1 = lim £ h h cos h + 1 h!0

a lim

h!0

cos2 h ¡ 1 h!0 h(cos h + 1)

= lim = lim

h!0

= lim

h!0

= lim

h!0

=1£ =0

¡ sin2 h h(cos h + 1) sin h ¡ sin h £ h cos h + 1 sin h ¡ sin h £ lim h h!0 cos h + 1

fby the limit laws since both limits existg

0 2

b Let f (x) = sin x. If f 0 (x) exists, then it is given by sin(x + h) ¡ sin x h sin x cos h + sin h cos x ¡ sin x = lim h h!0 h ³ ´ i cos h ¡ 1 sin h = lim sin x + £ cos x h h h!0 ³ ´³ ´ ³ ´³ ´ cos h ¡ 1 sin h = lim sin x lim + lim cos x lim h h!0 h!0 h!0 h!0 h

f 0 (x) = lim

h!0

fby the limit laws, since each of these limits existsg = sin x £ 0 + cos x £ 1 fsince x is independent of hg = cos x

DIFFERENTIABILITY AND CONTINUITY

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is differentiable at x = a, then f is continuous at x = a.

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26

CALCULUS

lim f(a + h) ¡ f (a)

Proof:

h!0

f (a + h) ¡ f (a) £h h f (a + h) ¡ f (a) £ lim h = lim h h!0 h!0

= lim

h!0

fby the limit laws since both limits existg

= f 0 (a) £ 0 =0 )

lim f (a + h) = f (a)

h!0

By letting x = a + h, and since h can take both positive and negative values, this is equivalent to lim f (x) = f(a). x!a

) f is continuous at x = a. It follows that if a function is not continuous at x = a, then it is not differentiable at x = a. The converse of this theorem is not true, however. If a function is continuous at x = a, it is not necessarily differentiable there. The set of all differentiable functions is therefore a proper subset of the set of all continuous functions.

TESTING FOR DIFFERENTIABILITY For functions which are defined by different expressions on separate intervals, we need a formal test to see whether the function is differentiable. To do this we first need to define: f (a + h) ¡ f (a) h

0 (a) = lim ² the left-hand derivative of f at x = a is f¡

h!0¡

0 (a) = lim ² the right-hand derivative of f at x = a is f+

h!0+

f (a + h) ¡ f (a) . h

A function f : D ! R is differentiable at x = a, a 2 D, if: 1 f is continuous at x = a, and 0 (a) = lim 2 f¡

h!0¡

f (a + h) ¡ f (a) h

Example 7

½

Prove that f (x) = jxj =

h!0+

f (a + h) ¡ f (a) h

x!0¡

y

x!0¡

lim f (x) = lim x = 0

and

x!0+

x!0+

y = jxj

lim f (x) = f(0) = 0

x!0

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) f is continuous at x = 0.

5

both exist and are equal.

x, x>0 is continuous but not differentiable at x = 0. ¡x, x < 0

lim f (x) = lim (¡x) = 0,

f (0) = 0,

)

0 (a) = lim and f+

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CALCULUS

½ Now

f 0 (x)

=

1, x>0 ¡1, x < 0

27

since the derivative of f exists on the open interval ] ¡1, 0 [

and on the open inteval ] 0, 1 [ . )

0 0 (0) = ¡1 and f+ (0) = 1 f¡

)

0 0 f¡ (0) 6= f+ (0)

) f is not differentiable at x = 0.

Example 8

½

Consider f(x) =

sin x, x>0 with domain R . 2 x + 5x, x < 0

a Prove that f is continuous but not differentiable at x = 0. b Write down f 0 (x) as a piecewise defined function. a f (0) = 0 is defined.

y

lim f (x) = lim (x2 + 5x) = 0

x!0¡

and )

x!0¡

x

lim f (x) = lim sin x = 0

x!0+

x!0+

y = f(x)

lim f (x) = f(0) = 0

x!0

) f(x) is continuous at x = 0. Using known derivatives for open intevals of R , ½ cos x, x>0 we have f 0 (x) = 2x + 5, x < 0 0 (0) = lim (2x + 5) = 2 £ 0 + 5 = 5 f¡

)

x!0¡

0

and f+ (0) = lim cos x = cos 0 = 1 0 f¡ (0)

Since

x!0+

0 6= f+ (0), f is not differentiable at x = 0.

We observe on the graph of y = f (x) that the curve does not have a unique tangent at x = 0. ½ cos x, x>0 0 b From a we have f (x) = 2x + 5, x < 0.

DISCUSSION There are functions which are continuous everywhere in their domain but which are differentiable nowhere!

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Research and discuss the Weierstrass function.

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28

CALCULUS

EXERCISE D 1 Let f (x) = cos x. Use the definition of the derivative to prove that f 0 (x) = ¡ sin x. ½ x ¡ 5, x > 5 2 Prove that f(x) = j x ¡ 5 j = is continuous but not differentiable at x = 5. 5 ¡ x, x < 5 ½ y x + 2, x>0 3 Let f (x) = 2 x + 3x, x < 0. y = f(x) Explain why f (x) is not differentiable at x = 0.

x

-3

½ 4 Let f (x) =

¡x2 + 5x + 6, x > 1 3x + 10, x < 1.

a Sketch the function y = f (x). 0 (1) i f¡

b Calculate:

0 (1) ii f+

c Is f differentiable at x = 1? Explain your answer. 5 For each of the following functions and the given value of a, determine whether the function is differentiable at x = a. ½ ½ 1 + sin x, x>0 cos x, x > 0 a f (x) = b f (x) = , a=0 , a=0 x2 + x + 1, x < 0 x3 , x2 c f (x) = , a=2 3 x + 2x + 1, x < 2 6 Investigate the continuity and differentiability of f at x = 0 if ½ k sin x, x > 0 f(x) = , where k 2 R is any constant. tan x, x < 0 7 Find constants c, d 2 R so that the given function is differentiable at x = 1. ½ 2 ½ sin(x ¡ 1) + cx, x > 1 x , x61 a f (x) = b f (x) = x 1 x2 ¡ x + d, ³ ´ ½ 3 1 x sin , x 6= 0 8 Let f (x) = x 0, x = 0. a Prove that f is continuous and differentiable at x = 0. b Write down f 0 (x) as a piecewise defined function.

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CALCULUS

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ˆ L’HOPITAL’S RULE

The limit laws do not help us to deal with limits which have indeterminate forms. These include: Type

Description

0 0

lim

x!a

1 1

f (x) g(x)

lim

x!a

0£1

lim f (x) = 0 and

where

x!a

lim g(x) = 0

x!a

where f(x) ! §1 and g(x) ! §1 when x ! a

lim [f(x) g(x)] where

x!a

For example, consider

lim f(x) = 0 and g(x) ! §1 when x ! a

x!a

2x ¡ 1 . x!0 x

lim

lim (2x ¡ 1) = 0 and

Since

f (x) g(x)

lim (x) = 0, we have the indeterminate form

x!0

x!0

0 0.

To address these types of limits, we use l’Hôpital’s Rule: Suppose f (x) and g(x) are differentiable and g0 (x) 6= 0 on an open interval that contains the point x = a. If

lim f (x) = 0 and

lim g(x) = 0, or, if as x ! a, f (x) ! §1 and g(x) ! §1,

x!a

then

x!a

f (x) f 0 (x) = lim 0 x!a g(x) x!a g (x)

lim

provided the limit on the right exists.

Proof of one case of l’Hôpital’s Rule: The derivative of a function f (x) at a point x = a, denoted by f 0 (a), is given by f 0 (a) = lim

x!a

f (x) ¡ f (a) . x¡a

Using this definition of the derivative, we prove the case of l’Hôpital’s Rule in which f (a) = g(a) = 0, f 0 (x) and g0 (x) are continuous, and g 0 (a) 6= 0. Under these conditions, f (x) f (x) ¡ f (a) = lim x!a g(x) x!a g(x) ¡ g(a)

lim

f (x)¡f (a) x¡a x!a g(x)¡g(a) x¡a

fmultiplying by

= lim

f (x)¡f (a) x¡a

lim

g(x)¡g(a) x¡a

lim f 0 (x)

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f 0 (x) g 0 (x)

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=

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25

=

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x!a

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x!a

fsince f(a) = g(a) = 0g

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30

CALCULUS

Example 9 Use l’Hôpital’s Rule to evaluate: 2x ¡ 1 x!0 x

a lim

lim xe¡x

b

c

x!1

lim

x!1

ln x x

a lim (2x ¡ 1) = 0 and lim x = 0, so we can use l’Hôpital’s Rule if the resulting limit exists. x!0

) =

x!0

2x ¡ 1 lim x!0 x

2x ¡ 1 x

lim

x!0

d lim dx (2x ¡ 1)

x!0

Pp .

has indeterminate form

fl’Hôpital’s Ruleg

d lim dx (x)

x!0

=

lim 2x ln 2

x!0

lim 1

x!0

=

ln 2 1

= ln 2

lim xe¡x has

b As x ! 1, e¡x ! 0, so we can use l’Hôpital’s Rule.

x!1

lim xe¡x

)

indeterminate form 1 £ 0.

x!1

= lim

x!1

=

x ex

fconvert limit to a quotient with the form

lim d (x) x!1 dx lim d (ex ) x!1 dx

1 g 1


lim 1

x!1

=

lim (ex )

x!1

=0

fsince

lim 1 = 1 and as x ! 1, ex ! 1g

x!1

c As x ! 1, ln x ! 1 and x ! 1, so we can use l’Hôpital’s Rule. )

lim

x!1

=

=

ln x x

lim d (ln x) x!1 dx lim d (x) x!1 dx lim


ln x 1 . has indeterminate form x 1

³ ´ 1 x

x!1

lim 1

x!1

=

lim

x!1

0 1

fsince

lim

³ ´ 1 x

x!1

=0

= 0g

Important: A common error is to attempt to evaluate the Fundamental Trigonometric Limit lim

x!0

sin x x

using l’Hôpital’s Rule. We cannot use l’Hôpital’s Rule in this case as the derivative of sin x from first principles itself requires the use of

lim

x!0

sin x = 1, x

as shown in

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Example 6.

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CALCULUS

EXERCISE E 1 Evaluate, if possible, the following limits using l’Hôpital’s Rule. 1 ¡ cos x x!0 x2

b

ex ¡ 1 ¡ x x!0 x2

ln x x!1 x ¡ 1

d

ex x!1 x

a lim c lim e

lim x ln x

f

x2 + x x!0 sin 2x

h

x!0+

g lim i lim

x + sin x x ¡ sin x

k lim

ax ¡ bx , sin x

x!0 x!0

2 Attempt to find

lim

3 Show that

lim

lim

lim

x!0

Some of these limits were evaluated in Exercises B.2 and B.3 using other methods.

arctan x x

lim

x!0+

sin x p x

lim x2 ln x

j

x!0+

where a, b > 0 are real constants. lim

¼¡ x! 2

¼ 2

tan x sec x

using l’Hôpital’s Rule.

¡ arccos x ¡ x

= 16 .

x3

x!0

Example 10 Find

lim

x!0+

ln(cos 3x) . ln(cos 2x)

lim ln(cos 3x) = 0 and

x!0+

ln(cos 3x) ) lim = lim x!0+ ln(cos 2x) x!0+

lim ln(cos 2x) = 0, so we can use l’Hôpital’s Rule.

x!0+

Ã ¡3 sin 3x ! cos 3x


¡2 sin 2x

³

cos 2x

´ = lim x!0+ µ ¶ µ ¶ sin 3x 3 cos 2x = lim £ lim 3 sin 3x cos 2x 2 sin 2x cos 3x

x!0+

sin 2x

lim

sin 3x sin 2x

µ

lim sin 3x = 0 and

£

3 2

lim sin 2x = 0,

x!0+

x!0+

so we can use l’Hôpital’s Rule again. µ ¶ ln(cos 3x) 3 cos 3x ) lim = lim £ x!0+

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=

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=

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100

ln(cos 2x)

25

x!0+

At step (¤) we could have alternatively used the Fundamental Trigonometric Limit.

(¤)

75

Now

x!0+

25

=

2 cos 3x

x!0+

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31

32

CALCULUS

4 Evaluate: ln(cos 5x) ln(cos 3x)

lim

a

x!0+

lim

b

¼¡ x! 2

ln(sin 2x) ln(sin 3x)

Example 11 lim (sec x ¡ tan x).

Evaluate

¼¡

x! 2

As x !

¼¡ , 2

both sec x ! 1 and tan x ! 1.

We therefore need to convert the difference sec x ¡ tan x into a quotient, and then apply l’Hôpital’s Rule. Now sec x ¡ tan x = )

1 sin x 1 ¡ sin x ¡ = cos x cos x cos x

³

lim (sec x ¡ tan x) = lim ¼¡

¼¡

x! 2

x! 2

1 ¡ sin x cos x

´

lim (1 ¡ sin x) = 0 and

where

lim cos x = 0

¼¡

¼¡

x! 2

lim

)

lim (sec x ¡ tan x) = ¼¡

x! 2

x!

¼¡ 2

lim

¼¡ x! 2

=

0 1

x

(¡ cos x)


(¡ sin x)

=0

5 Evaluate, if possible: ³ ´ 1 1 a lim ¡ x!0+

x! 2

b

sin x

lim

³

x!0+

1 1 ¡ x sin 2x

´ c

lim (sec2 x ¡ tan x) ¼¡

x! 2

Example 12 ex x!1 xn

where n 2 Z + .

lim

Evaluate, if possible:

For all n 2 Z + , as x ! 1, ex ! 1 and xn ! 1, so we can use l’Hôpital’s Rule. )

ex ex = lim n x!1 x x!1 nxn¡1 ex = lim x!1 n(n ¡ 1)xn¡2

lim

We use l’Hôpital’s Rule n times.

.. . ex x!1 n! 1 = lim ex n! x!1

= lim

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As x ! 1, ex ! 1, so the given limit DNE.

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CALCULUS

6

xk , x!1 ex

k 2 Z +.

lim

a Evaluate, if possible:

33

b Hence explain why, as x ! 1, the exponential function ex increases more rapidly than any fixed positive power of x. 7 Prove that as x ! 1, ln x increases more slowly than any fixed positive power of x. ³ ´ 1 8 a Prove that lim x ln 1 + = 1. x!1 x ¡ ¢ ³ ´ 1 1 x x ln 1+ x b By writing 1+ =e and using the fact that f (x) = ex is continuous on R , x ³ ´ 1 x = e. prove that lim 1 + x!1 x ³ ´ a x c Prove that for a 6= 0, lim 1 + = ea . x!1

x

9 By writing xsin x = esin x ln x and using the fact that f (x) = ln x is continuous on x 2 R , x > 0, prove that lim xsin x = 1. x!0+

1

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10 Prove that

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34

CALCULUS

F

ROLLE’S THEOREM AND THE MEAN VALUE THEOREM (MVT)

Suppose a real function f is defined on domain D. f : D ! R is continuous on a closed interval [a, b], a < b, if: 1 f is continuous at x = c for all c 2 ] a, b [ , and 2

lim f(x) = f (a) and

x!a+

lim f (x) = f (b)

x!b¡

For a function f : D ! R we define: f is differentiable on an open interval ] a, b [ , a < b if f is differentiable at x = c for all c 2 ] a, b [ . f is differentiable on a closed interval [a, b] , a < b, if: 1 f is differentiable on ] a, b [ , and 0 (a) and f 0 (b) both exist. 2 f+ ¡

ROLLE’S THEOREM Suppose function f : D ! R is continuous on the closed interval [a, b], and differentiable on the open interval ] a, b [.

y y = f(x)

If f (a) = f (b) = 0, then there exists a value c 2 ] a, b [ such that f 0 (c) = 0.

a

c

b

x

Rolle’s theorem guarantees that between any two zeros of a differentiable function f there is at least one point at which the tangent line to the graph y = f(x) is horizontal. Proof of Rolle’s theorem: Since f is continuous on [a, b], it attains both a maximum and minimum value on [a, b]. If f takes positive values on [a, b], let f(c) be the maximum of these. Now f (a) = 0 = f(b) and f (c) > 0, so c 2 ] a, b [ . Since f is differentiable at c, f must have a local maximum at x = c. ) f 0 (c) = 0. Similarly, if f takes negative values on [a, b], let the minimum of these be f (c). It follows that f has a local minimum at x = c, and therefore f 0 (c) = 0. Finally, if f (x) = 0 for all x 2 [a, b] then clearly f 0 (c) = 0 for all c 2 ] a, b [ . Rolle’s theorem is a lemma used to prove the following theorem.

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A lemma is a proven proposition which leads on to a larger result.

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CALCULUS

35

THE MEAN VALUE THEOREM (MVT) (THE LAGRANGE FORM) If f is a function continuous on [a, b] and differentiable on ] a, b [, then f (b) ¡ f (a) = f 0 (c) £ (b ¡ a) for some number c 2 ] a, b [ . The MVT tells us that for such a function f there exists at least one value c 2 ] a, b [ such that the tangent to the curve y = f (x) at x = c has gradient equal to the gradient of the chord through points (a, f(a)) and (b, f (b)).

y

y = f(x) gradient = f(b) ¡ f(a) b¡a gradient = f 0 (c)

a

c

b

x

Proof of the MVT: Let h(x) = f (x) ¡

h

i

f (b) ¡ f (a) (x ¡ a) ¡ f (a) b¡a

for x 2 [a, b].

Since f is continuous on [a, b] and differentiable on ] a, b [ , so is the function h. Now, h(a) = h(b) = 0, so by Rolle’s theorem there exists c 2 ] a, b [ such that h0 (c) = 0. ³ ´ f (b) ¡ f (a) for x 2 ] a, b [ But h0 (x) = f 0 (x) ¡ b¡a ³ ´ f (b) ¡ f (a) ) h0 (c) = f 0 (c) ¡ =0 b¡a

f (b) ¡ f (a) . ) f 0 (c) = b¡a

We have seen that the derivative of a constant function is zero. We can now prove the converse of this result, that if a continuous function f has derivative equal to zero, then f is a constant function. FIRST COROLLARY OF THE MVT Suppose the function f is continuous on [a, b] and differentiable on ] a, b [ . If f 0 (x) = 0 for all x 2 ] a, b [ , then f (x) is a constant function on [a, b]. Proof: Let x 2 ] a, b ] and apply the MVT to f on the interval [a; x].

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Then f (x) ¡ f (a) = f 0 (c)(x ¡ a) for some c 2 ] a, x [ = 0 £ (x ¡ a) =0 ) f (x) = f (a) for all x 2 ] a, b ] ) f (x) = f (a) for all x 2 [a, b]

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A corollary is a result which follows directly from a theorem, and which is worth stating in its own right.



36

CALCULUS

SECOND COROLLARY OF THE MVT Suppose F and G are two functions continuous on [a, b] and differentiable on ] a, b [ . If F 0 (x) = G0 (x) for all x 2 ] a, b [ , then F (x) = G(x) + C for some constant C, for all x 2 [a, b]. Let H(x) = F (x) ¡ G(x), for x 2 [a, b].

Proof:

) H is differentiable on ] a, b [ and H 0 (x) = 0 on ] a, b [ . By the first corollary of the MVT, H(x) is a constant for all x 2 [a, b]. Hence F (x) = G(x) + C

as required.

ANTIDERIVATIVES A function f has an antiderivative G on [a, b] if there exists a function G continuous on [a, b] such that G0 (x) = f (x) for all x 2 ] a, b [ . If a function f has an antiderivative G, then G is unique up to the addition of a constant. Suppose f has two antiderivatives F , G on [a, b].

Proof:

By definition, F 0 (x) = G0 (x) = f (x) for all x 2 ] a, b [ . By the second corollary to the MVT, F (x) = G(x) + C, where C is a constant.

Z f (x) dx = G(x)+C

We define

to be the indefinite integral of function f with respect to x. The

function f is called the integrand and the indefinite integral of f is thus the set of all antiderivatives of f on [a, b].

EXERCISE F 1 Determine whether or not Rolle’s theorem applies to the function f on the given interval [a, b]. If Rolle’s theorem does apply, find all values c 2 ] a, b [ for which f 0 (c) = 0. a f (x) = 3x3 + 5x2 ¡ 43x + 35, [a, b] = [¡5, 2 13 ] b f (x) = jxj ¡ 5, [a, b] = [¡5, 5] 1

c f (x) = 2 ¡ , [a, b] = [¡ 12 , 7] x+1 ½ ¡2x ¡ 5, x < ¡1 d f (x) = , [a, b] = [¡2 12 , 2] x > ¡1 x2 ¡ 4, 2 For each of the following functions, find: i the number of real zeros of the derivative f 0 (x), as guaranteed by Rolle’s theorem ii the exact number of real zeros of the derivative f 0 (x). b f (x) = (x ¡ 1)2 (x2 ¡ 9)(x ¡ 2)

a f (x) = (x ¡ 1)(x ¡ 2)(x ¡ 4)(x ¡ 5)

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c f (x) = (x ¡ 1)2 (x2 + 9)(x ¡ 2)

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CALCULUS

37

3 Show that the given function f satisfies the MVT on the given interval [a, b]. Find all values of c such that f(b) ¡ f (a) = f 0 (c) £ (b ¡ a). p a f(x) = x3 , [a, b] = [¡2, 2] b f(x) = x ¡ 2, [a, b] = [3, 6] 1 x

c f(x) = x + , [a, b] = [1, 3] 4 Let f (x) =

p x.

a Use the MVT to show there exists c 2 ] 49, 51 [ such that

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c Hence prove

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b Graph

1 y= p x

p 1 51 ¡ 7 = p .

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38

CALCULUS

G

RIEMANN SUMS

Let y = f (x) be a function which is non-negative and continuous on the interval [a, b], a < b. y y = f(x) A

a

x

b

Let A denote the area under the curve y = f (x) and above the x-axis for a 6 x 6 b. Archimedes of Syracuse (c. 287 - 212 BC) developed the following method for approximating the area A using the sum of areas of rectangles: A partition of the interval [a, b] is a set of n + 1 points in [a, b] which divide the interval into n subintervals [x0 , x1 ], [x1 , x2 ], ...., [xn¡1 , xn ], where x0 = a, xn = b, and x0 < x1 < :::: < xn¡1 < xn . We write partition P = fa = x0 , x1 , ...., xn = bg. The length of the ith subinterval [xi¡1 , xi ], i = 1, ...., n is ¢xi = xi ¡ xi¡1 . y y = f(x)

a = x0

xn = b x

xn-1

x2

x1

Let xi¤ 2 [xi¡1 , xi ] be an arbitrarily chosen value in the ith subinterval, i = 1, ...., n. Let Ai = the area of the rectangle whose base is the width of the ith subinterval and with height equal to f (xi¤ ) ) Ai = f (xi¤ ) £ ¢xi y y = f(x)

f(x2*) A2 x0

x1 x2* x2

The area A can be approximated by A ¼

n P i=1

xn

x

Ai = f(x1¤ )¢x1 + :::: + f (xn¤ )¢xn =

n P

f (xi¤ )¢xi

i=1

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and this is called a Riemann sum.

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CALCULUS

39

LOWER AND UPPER RIEMANN SUMS Suppose f is non-negative and continuous on [a, b], and let partition P = fa = x0 , x1 , ...., xn = bg. If we choose the values xi¤ 2 [xi¡1 , xi ], i = 1, ...., n so that f (xi¤ ) is the minimum value of f (x) on the interval [xi¡1 , xi ], then the Riemann Sum is called a lower Riemann sum and is denoted n P f (xi¤ )¢xi . Ln = i=1

If we choose the values xi¤ 2 [xi¡1 , xi ], i = 1, ...., n so that f(xi¤ ) is the maximum value of f (x) on the interval [xi¡1 , xi ], then the Riemann Sum is called a upper Riemann sum and is denoted n P Un = f (xi¤ )¢xi . i=1

REGULAR PARTITIONS Partition P is regular if the n subintervals have equal length. In this case: b¡a = ¢xi n³ ´ b¡a i ² xi = a + n

² ¢x =

for all i = 1, ...., n

for all i = 0, ...., n ³ ´ n b¡a P f (xi¤ ). ² the Riemann sum is n

i=1

When evaluating Riemann sums, the following identities may be useful: ² ² ²

n P i=1 n P i=1 n P

c = cn for c a constant i=

n(n + 1) 2

i2 =

i=1

n(n + 1)(2n + 1) 2n3 + 3n2 + n = 6 6

CALCULATING EXACT AREAS For the function f(x), the rectangles for the lower sum lie under the curve y = f (x) and the rectangles for the upper sum lie above the curve. The exact area A under the curve and above the x-axis on [a, b] therefore satisfies Ln 6 A 6 Un for all partitions of [a, b] into n subintervals, n 2 Z + . For a regular partition, in the limit as n ! 1, the width of each subinterval ! 0. lim ¢x = lim

n!1

n!1

b¡a = 0. n

At the same time, the approximation for the area A is squeezed between the lower and upper Riemann sums. lim Ln = lim A = lim Un n!1

n!1

n!1

For f a non-negative continuous function on [a, b], a < b, the exact area A under the curve y = f (x) and above the x-axis on [a, b] is the unique value which satisfies Ln 6 A 6 Un for all regular partitions of [a, b].

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lim Ln = A = lim Un .

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In particular,

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40

CALCULUS

Example 13 Suppose f (x) = x2 on [0, 1]. a Let P be a regular partition of [0, 1] into n subintervals. Find ¢x, and write xi in terms of i and n. 1 1 1 ¡ + 2. 3 2n 6n 1 1 1 Un = + + 2. 3 2n 6n

b Show that the lower Riemann sum is given by Ln = c Show that the upper Riemann sum is given by

d Hence find the area A under the curve y = x2 and above the x-axis on [0, 1]. a There are n subintervals from x0 = 0 to xn = 1.

y

1¡0 1 = ) ¢x = n n

y = x2

) xi = x0 + i¢x =

i , n

M3

i = 1, ...., n.

m3 1 n

2 n

3 n

x

1

b The minimum value of f(x) on [xi¡1 , xi ] c The maximum value of f (x) on [xi¡1 , xi ] ³ ´ ³ ´2 i¡1 2 i , i = 1, ...., n. , i = 1, ...., n. is mi = f (xi¡1 ) = is Mi = f (xi ) = n

n

) the lower Riemann sum is n n 1 P 1 P (i ¡ 1)2 Ln = mi = 2 n i=1

n i=1

=

n 1 P

n3

) the upper Riemann sum is n n 1 P 1 P i2 Mi = Un = 2

n

n i=1

(i ¡ 1)2

=

i=1 nP ¡1

1 i2 n3 i=0 ³ ´ 1 (n ¡ 1)(n)(2n ¡ 1) = 3 n 6

=

1 = 3 n

=

n i=1 n

µ

2n3 ¡ 3n2 + n 6

n 1 P i2 3 n i=1

1 = 3 n

=

¶

µ

2n3 + 3n2 + n 6

¶

1 1 1 + + 2 3 2n 6n

1 1 1 ¡ + 2 3 2n 6n

d The rectangles for the lower sum lie under the curve y = x2 and the rectangles for the upper sum lie above the curve. ) the exact area A under the curve y = x2 and above the x-axis on [0, 1] satisfies Ln 6 A 6 Un for all regular partitions of [0, 1] into n subintervals, n 2 Z + . 1 1 1 1 1 1 ¡ + 2 6A6 + + 2 for all n 2 Z + . 3 2n 6n 3 2n 6n ³ ´ ³ ´ 1 1 1 1 1 1 1 Now lim ¡ + 2 = = lim + + 2 n!1 n!1 3 2n 6n 3 3 2n 6n

)

lim A = A, since A is a constant.

and

n!1

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41

EXERCISE G.1 1 Let y = sin x on [0, ¼]. Let P = f0, ¼4 , ¼2 , 3¼ 4 , ¼g be a regular partition of [0, ¼]. Calculate L4 and U4 exactly, and sketch the corresponding areas on graphs of the function. 2 Suppose f (x) = 2x on [1, 4]. a Let P be a regular partition of [1, 4] into n subintervals. Find ¢x, and write xi in terms of i and n. b Write expressions for Ln and Un . c Hence find the area A under the curve f (x) = 2x and above the x-axis on [1, 4]. 3 Suppose f (x) = x2 on [1, 2]. a Let P be a regular partition of [1, 2] into n subintervals. Find ¢x, and write xi in terms of i and n. b Write expressions for Ln and Un . c Hence show that the area A under the curve f (x) = x2 and above the x-axis on [1, 2] is 7 2 3 units .

DEFINITE INTEGRALS Let f , not necessarily non-negative or continuous, be any function defined on the interval [a, b]. Let P = fa = x0 , x1 , ...., xn = bg be any partition of [a, b]. If the values xi¤ 2 [xi¡1 , xi ], i = 1, ...., n are arbitrarily chosen then R =

n P

f (xi¤ )¢xi

where

i=1

¢xi = xi ¡ xi¡1 , i = 1, ...., n is called the Riemann sum of f on [a, b] for partition P and selection fx1¤ , x2¤ , ...., xn¤ g. b¡a = ¢xi n

If P is a regular partition then ¢x =

In the limit as n ! 1, ¢x ! 0 and

for all i = 1, ...., n, and R =

n (b ¡ a) P f (xi¤ ). n i=1

n (b ¡ a) P f(xi¤ ). n!1 n i=1

lim R = lim

n!1

If this limit exists, we say f is integrable on [a, b]. We denote the value of this limit by Z b n (b ¡ a) P f (x) dx = lim f (xi ) and call this the definite integral of function f from a to b. n!1

a

n

i=1

The function f is the integrand. If a function f is continuous on [a, b], then f is integrable on [a, b].

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The interval [a, b] is called the domain of integration. The values a and b are also called, respectively, the lower and upper limits of the integral.

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42

CALCULUS

INTERPRETING DEFINITE INTEGRALS AS AREAS If f is an integrable function which takes negative values on [a, b], we must take care when interpreting definite integrals as areas. We observe that: ² If f(x) > 0 on [a, b], a < b, then Z b f (x) dx = A = the area under the curve.

y

y = f(x)

a

A

² If f(x) 6 0 on [a, b] with a < b, then Z b f (x) dx = ¡A = ¡(the area between the curve a and the x-axis) In this case the values f (xi¤ ) in the corresponding Riemann sum will be negative.

Z

a

b

a

b

y

y = f(x)

Z

b

a

function on [a, b] with a < b.

x

A

f (x) dx = ¡

If we reverse the direction of integration we obtain

x

a

f (x) dx for f an integrable b

By interpreting definite integrals in terms of areas, the following results can be demonstrated: For f(x), g(x) integrable functions on [a, b]: ² If f takes positive and negative values on [a, b] then the area bounded by y = f(x), the x-axis, and the lines x = a and x = b is given by Z b jf(x)j dx.

y

y = f(x)

a

b

x

a

In this case, Z b f(x) dx = (enclosed area above the x-axis) a

a

y

f(x) dx = 0

Z

c

f (x) dx + a

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x

g(x) dx

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f (x) dx§

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Z

b

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for all c 2 [a, b] Z b Z (f(x)§g(x)) dx = ² a

f(x) dx c

0

a

y = f(x)

b

5

f (x) dx =

95

Z

b

25

Z ²

100

a

5

²

75

Z

¡ (enclosed area below the x-axis)

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CALCULUS

EXERCISE G.2

Z

1 Suppose f is a function continuous on R , and that Z 5 Z 6 1 f(x) dx = ¡ 2 , and f (x) dx = ¡2. 2

3

Z

7

f (x) dx = 2,

1

2

43

f (x) dx = 12 ,

3

a Find: Z i

6

Z f (x) dx

Z

5

f (x) dx

ii

1

v

Z

7

Z f (x) dx ¡

3

5

iii

5

Z f (x) dx

7

Z

2

vi

f(x) dx 1

2

3

4

iv

Z f (x) dx +

f (x) dx

4 7

f(x) dx 6

b Sketch a possible graph for y = f(x) on [1, 7]. 8 06x 0 be such that x + h < b.

y

y = f(t)

x x+h

a

Now

F (x + h) ¡ F (x) 1 = h h

=

1 h

µZ

Z

x+h

x

f(t) dt ¡ a x+h

Z

b

t

¶ f (t) dt

a

f (t) dt x

1 = (x + h) ¡ x

Z

x+h

f (t) dt x

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= f (c) for some c 2 [x, x + h], since f is continuous on [a, b].

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46

CALCULUS

)

lim

h!0+

F (x + h) ¡ F (x) = lim f(c) = f(x) h h!0+

since x 6 c 6 x + h and f is continuous on [a, b].

In the same way for h < 0, x + h > a, we find lim

h!0¡

F (x + h) ¡ F (x) = lim f(c) = f(x) h h!0¡

since x + h 6 c 6 x and f is continuous on [a, b].

) F 0 (x) = f(x) as required. Since F is differentiable on ] a, b [ , F is continuous on ] a, b [ . We still need to show that F is continuous on [a, b]. Since f is continuous on [a, b], f is integrable on [a, b] Z a Z b ) F (a) = f (t) dt = 0 and F (b) = f(t) dt a

Now

Z

are both defined.

a

Z

x

f (t) dt lim F (x) = lim x!a+ x!a+ a Z a f (t) dt =

and

lim F (x) = lim

x!b¡

x!b¡

Z

f(t) dt a

b

f (t) dt

=

a

x

a

=0

= F (b).

= F (a) Thus F (x) is continuous on [a, b]. The FTOC (Part 1) can also be written as:

If f is continuous on [a, b] then

d dx

µZ

x

¶ f (t) dt = f(x), for x 2 ] a, b [ .

a

The FTOC (Part 1) guarantees that any function f continuous on [a, b] has the antiderivative Z x f(t) dt on [a, b], even if we cannot write it down. F (x) = a

Z

x

For example, f (x) = sin(x3 ) has an antiderivative F (x) =

sin(t3 ) dt on [ 0, 1 [ even though

0

we cannot write down an antiderivative of sin(x3 ) explicitly as a function.

THE FUNDAMENTAL THEOREM OF CALCULUS (PART 2) If f is any function integrable on [a, b] which has an antiderivative G on [a, b], Z b f (x) dx = [G(x)]ba = G(b) ¡ G(a).

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CALCULUS

47

Proof: Case: If f is continuous on [a, b] then f is integrable on [a, b]. Z x By FTOC (Part 1) the function F (x) = f(t) dt is an antiderivative of f on [a, b]. a

By the second corollary to the MVT, F (x) = G(x) + C ) F (a) = G(a) + C

Z

where C is a constant.

a

f (t) dt = G(a) + C

) a

) 0 = G(a) + C ) C = ¡G(a).

Hence F (x) = G(x) ¡ G(a) ) F (b) = G(b) ¡ G(a) Z b f(t) dt = G(b) ¡ G(a) as required. ) a

Z

Case: Suppose f is not continuous on [a, b]. By the premise of the theorem,

b

f(x) dx a

exists, and G is a function continuous on [a, b] such that G0 (x) = f (x) for x 2 ] a, b [ . Let P = fa = x0 , x1 , ...., xn = bg be a regular partition of [a, b] with n 2 Z + . Consider the ith subinterval [xi¡1 , xi ], i = 1, 2, ...., n. By the Mean Value Theorem,

G(xi ) ¡ G(xi¡1 ) = G0 (xi¤ ) xi ¡ xi¡1

Thus G(xi ) ¡ G(xi¡1 ) = G0 (xi¤ )¢x, where ¢x =

for some xi¤ 2 [xi¡1 , xi ].

b¡a = xi ¡ xi¡1 . n

Summing over all subintervals in the partition we obtain n P

G0 (xi¤ )¢x =

i=1

n P

(G(xi ) ¡ G(xi¡1 ))

i=1

= [G(x1 ) ¡ G(x0 )] + [G(x2 ) ¡ G(x1 )] + :::: + [G(xn ) ¡ G(xn¡1 )] = G(xn ) ¡ G(x0 ) )

n P

f(xi¤ )¢x = G(b) ¡ G(a), since G0 (xi¤ ) = f(xi¤ ), xn = b, and x0 = a.

i=1

)

·

lim

n!1

n P

¸ f(xi¤ )¢x = G(b) ¡ G(a) which is a constant.

i=1

Z

b

)

f (x) dx = G(b) ¡ G(a) as required. a

If a function is integrable and has an explicitly defined antiderivative, then the FTOC (Part 2) allows us to calculate definite integrals easily. · ¸1 Z 1 3 3 x3 x3 and therefore x2 dx = = 13 ¡ 03 = 13 . For example, if f (x) = x2 then G(x) =

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48

CALCULUS

DISCUSSION Does the existence of an antiderivative guarantee that a function defined on [a, b] is integrable on [a, b]?

Example 14 Find

F 0 (x)

Z

x

a F (x) =

if:

Z

3

cos t dt

x

¶ cos t dt

µZ

¶ et dt

5

x2

Ã Z d = ¡

3

1

dx

3

= cos x

d = dx

This is valid since f(t) = cos3 t is a continuous function.

d = du

!

x2

et dt

5

Ã Z ¡ µ Z ¡

u

¶ du e dt £ t

dx

5

= ¡2xex

Z

x2

if F (x) = x

d = du

c

ÃZ

u

µZ

c u

1 dt t2 + 3

d + dx

1 1 £ 2x ¡ 2 u2 + 3 x +3 2x 1 ¡ 2 = 4 x +3 x +3

95

50

c

¶

1 dt t2 + 3

fletting u(x) = x2 g fChain ruleg

fvalid since f (t) =

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where c is a constant

x +3

dx

=

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µ Z ¡

¶ 1 1 dt ¡ 2 t2 + 3 x +3 ¶ 1 du 1 dt £ ¡ 2 2 t +3

c

1 dt t2 + 3

!

x2

c

µZ

!

c

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d dx

1 dt + t2 + 3 x

75

=

Z

x2

25

d = dx

1 dt t2 + 3

x

ÃZ

fvalid since f(t) = et is continuousg

2

!

x2

0

d = dx

ÃZ

5

d F 0 (x) = dx

fChain ruleg

1 dt. t2 + 3

100

Find

F 0 (x)

where u(x) = x2

et dt

5

= ¡eu £ 2x

Example 15

!

u(x)

95

d = dx

d dx

b F 0 (x) =

1 is continuousg t2 + 3

100

F 0 (x) µZ

et dt x2

1

a

5

b F (x) =

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49

CALCULUS

EXERCISE H 1 Find F 0 (x) if: Z x a F (x) = sin4 t dt

Z

1

Z

dt (t2 + 2)15

0

4

c F (x) =

x

b F (x) =

t2

e

Z

x2

esin t dt

d F (x) =

dt

x

¡1

Z

2

1 dx = 2 x ¡1

2 Explain what is wrong with the argument:

h

i2

¡1 x ¡1

fFTOCg

= ¡ 12 ¡ (1) 3

d dx

a Can

µZ

x

= ¡ 32

¶ 1 dt for x 2 R

be calculated using the Fundamental Theorem of Calculus?

¡1 t

Explain your answer. b For which values of x can the derivative in a be calculated? 4 Calculate the following derivatives: ÃZ ! 2x d t a dt b p dx

1 + t4

1

Z

10

5 Find ¡3

Z

5

6 Find 0

d dx

ÃZ

!

x2

sin(e ) dt t

d dx

c

x

µZ

sin x

x3

¶

1 dt 1 + t2

j x j dx.

8 sin x, > < 2x, f(x) dx for f (x) = > : 1, x

0 6 x 6 ¼2 <x63

¼ 2

3<x65

7 Find exactly the average value of each function on the given interval: ½ 2 x , ¡2 6 x 6 1 1 a f(x) = 3 on [2, 5] b f (x) = on [¡2, 5] x 1<x65 ex , Z x 2 8 Let F (x) = cos(et ) dt, x > 1. Find exactly: 1 ³q ¡ ¢´ 0 a F (x) b F 0 (0) c F0 ln ¼4 d F 00 (x) 9 Let y = f (t), t 2 [0, 4] be the function with the Z x f (t) dt, graph shown. Let g(x) =

y

a g(1)

b g(3)

d g0 (3)

e g00 (3).

y = f(t)

5

0

x 2 [0, 4]. Find exactly:

e F 00 (0)

3

c g0 (1)

(3, 1)

1 1

10 Suppose g 0 (3) = 5, g(3) = 4, f(2) = ¡1, and F (x) =

2

Z

3

semi circle

4

t

g(x)

f (t ¡ 2) dt. 1

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50

CALCULUS

I

IMPROPER INTEGRALS OF THE FORM Z 1 f (x) dx a

An integral is described as improper if one of the following is true: ² the integrand approaches +1 or ¡1 for one or more points on the domain of integration, Z 2 1 1 dx is improper since 2 is discontinuous at x = 0 and as x ! 0, f (x) ! 1 for example 2 ¡1 x

x

Z

Z

1

² it has the form

Z

b

f (x) dx, or

f(x) dx,

¡1

¡1

a

constants. We define, when these limits exist: Z 1 Z f (x) dx = lim b!1

a b

Z

and

f (x) dx

Z

b

f(x) dx a

Z

a

f (x) dx =

f (x) dx +

¡1

are any

a

a!¡1

Z

f(x) dx where a, b 2 R

b

f (x) dx = lim ¡1 Z 1

1

¡1

1

f (x) dx.

a

Z

In this course we are only concerned with improper integrals of the form

1

f (x) dx.

a

Z

Z

1

The improper integral

b

f(x) dx is said to be convergent if a

Z

1

where a 6 b < 1, and if

Z

b

f (x) dx = lim

b!1

a

f(x) dx exists for all b a

f (x) dx exists. a

Otherwise the improper integral is divergent. Example 16

Z

1

1 dx xp

Investigate the convergence of 1

Z If p = 1,

1

1

1 dx = lim xp b!1

Z

b

1

where p 2 R .

1 dx x

= lim [ln x]b1 b!1

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1 dx xp

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b!1

·

b

25

= lim

Z

0

1

1 dx = lim xp b!1

5

1

95

Z If p 6= 1,

x

1

100

Hence

= lim (ln b) which DNE since as b ! 1, ln b ! 1 b!1 Z 1 1 dx is divergent.

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CALCULUS

Z )

1

1 dx = lim xp b!1

1

= =

lim

Z

1

1

1 lim 1 ¡ p b!1 1 lim 1 ¡ p b!1

³ ´p¡1 1 b

b!1

Hence

1 (1 ¡ p)xp¡1

1 lim 1 ¡ p b!1

If p > 1 then If p < 1,

·

1 dx xp

¸b 1

·³ ´p¡1 ¸ b 1 x

·³ ´p¡1 1 b

·³ ´p¡1 1 b

1

¸ ¡1 .

¸ ¡1 =

1 . p¡1

DNE as b ! 1, (b)1¡p ! 1.

converges if p > 1 and diverges if p 6 1.

From Example 16 we conclude the important result:

Z

1

1 dx xp

1


THE COMPARISON TEST FOR IMPROPER INTEGRALS Suppose 0 6 f (x) 6 g(x) for all x > a. Z 1 Z g(x) dx is convergent, then so is ² If a

Z

1

² If

Z f (x) dx is divergent, then so is

a

Z

1

Determine whether 2

Now we know that 0 6

1 dx p x¡1

1

2

1 x¡1

1

Z

2

1 1 p dx ¡ p dx x x 1

Z

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Z

1

2

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is divergent, and so

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1 p dx x

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Z )

1

for all x > 2.

2

1 p dx x 1

where

Z

5

Z

1 p dx = x

g(x) dx.

p p x ¡ 1 6 x for all x > 2 1 x

Z

1

is convergent or divergent.

so it follows that 0 6 p 6 p Now

f(x) dx.

a

a

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Example 17

1

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1 p dx x

is divergent.

is divergent by the Comparison Test.



51

52

CALCULUS

Z

1

If

Z

1

j f (x) j dx converges then

a

f (x) dx converges.

a

Proof: By definition, ¡ j f (x) j 6 f(x) 6 j f(x) j ) 0 6 f(x) + j f(x) j 6 2 j f (x) j Z 1 Z f (x) + j f (x) j dx 6 2 ) 06 a

Z

1

a

Z

1

then

Z

1

f(x) + j f (x) j dx.

a

Z

1

j f (x) j dx = A and

Supposing

Z

j f (x) j dx is convergent then so is

a

1

j f(x) j dx

a

) by the Comparison Test, if

Z

1

f(x) + j f(x) j dx = B

where A, B 2 R

a

f (x) dx = B ¡ A.

a

1

Hence

f (x) dx is convergent.

a

Example 18

Z

1

Using integration by parts and the Comparison Test, prove that 1

Z

1

sin x dx = lim x b!1

1

Z

b

sin x dx x

is convergent.

sin x dx x

1

Z b h i cos x b cos x = lim ¡ ¡ lim dx fintegration by partsg x x2 b!1 b !1 1 1 Z ³ ´ 1 cos b cos x + cos 1 ¡ dx = lim ¡ b x2 b!1 1 Z 1 cos x dx = cos 1 ¡ 2

¯ ¯ 1 ¯ cos x ¯ Now 0 6 ¯ 2 ¯ 6 2 x

x

1

x

for all x > 1,

Z

1

1

dx is convergent. and we also know from Example 16 that x2 1 Z 1¯ Z 1 ¯ cos x ¯ cos x ¯ dx. ) ¯ 2 ¯ dx is also convergent, and hence so is 2

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CALCULUS

53

EVALUATING IMPROPER INTEGRALS When an improper integral is convergent, we may be able to evaluate it using a variety of techniques. These include use of the limit laws, l’Hôpital’s Rule, integration by parts, and integration by substitution. Example 19 Z Evaluate

1

xe¡x dx, for any constant a 2 R .

a

Z

1

Z

¡x

b

dx = lim

xe

b!1

a

xe¡x dx

a

µ ¶ Z b £ ¤b ¡xe¡x a ¡ ¡e¡x dx b!1 a ³ £ ¤ b´ ¡b ¡a = lim ¡be + ae ¡ e¡x a b!1 ¡ ¢ = lim ¡be¡b + ae¡a ¡ e¡b + e¡a b!1 ¡ ¢ = e¡a (a + 1) + lim e¡b (¡1 ¡ b) b!1 µ ¶ ¡1 ¡ b ¡a = e (a + 1) + lim b = lim

b!1

fintegration by partsg

e

Now as b ! 1, ¡1 ¡ b ! ¡1 and eb ! 1. Z 1 ¡1 ) xe¡x dx = e¡a (a + 1) + lim b fl’Hôpital’s Ruleg a

b!1

= e¡a (a + 1)

e

EXERCISE I.1 1 Use the Comparison Test for improper integrals to test for convergence: Z 1 Z 1 2 x x ¡1 a dx b dx p 5 2 2x + 3x + 1

1

Z

1

2 Determine whether 1

3 Test for convergence: Z 1 2 x +1 a dx 4 1

Z

1

4 Show that

sin x dx x3

Z

1

b

x +1

x7 + 1

2

is convergent.

Z

2

e¡x dx

1

c

0

1

Z

ln x dx x

1

d

e¡x ln x dx

1

e¡x cos x dx is convergent.

0

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25

1

6 Evaluate

0

Z

b

5

a

Z

dx x2 + a2

100

5 Evaluate: Z 1 a

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54

CALCULUS

Z

1

µ

7 Evaluate 1

¶

1 1 p ¡p x x+3

dx. 1

8 Find the area in the first quadrant under the curve y = 2 . x + 6x + 10 Z 1 ln x 9 Prove that dx is divergent for p 6 1. p x

e

10

Z

1

a Evaluate the integral

Z

xn e¡x dx for n = 0, 1, 2, 3.

0

1

xn e¡x dx for n 2 Z + .

b Predict the value of 0

c Prove your prediction using mathematical induction.

APPROXIMATION TO THE IMPROPER INTEGRAL

Z

1

a

f (x) dx, a 2 Z Z

Suppose f is continuous and positive on [a, 1 [ , a 2 Z . We consider the integral

1

f(x) dx in a

terms of area under the curve for x > a. Consider a regular partition P = fa, a + 1, a + 2, ....g of [ a, 1 [ with subintervals of length ¢x = 1. y

f(a)

f(a+1) f(a+2) f(a+3) ...

Z 1 Area = f (x)dx a

. e tc .

y = ¦(x)

a

x

a+1 a+2 a+3

For each interval of length one along the x-axis, we can draw a rectangle of height equal to the value of the function on one side of the rectangle. For example, using the left side of the rectangle, the rectangle from x = a to x = a + 1 would have height f (a), the rectangle from x = a + 1 to x = a + 2 would have height f (a + 1), and so on. y

f(a) f(a+1)

f(a+2) f(a+3) y = ¦(x)

a

x

a+1 a+2 a+3

Z 1 f (x) dx may As with definite integrals approximated by Riemann sums, the improper integral a be approximated by the corresponding sum of areas of these rectangles. Z 1 1 b 1 P P P f (x) dx ¼ f (i) = lim f (i) where f(i) is also called an infinite series.

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CALCULUS

55

DECREASING AND INCREASING FUNCTIONS Suppose the function f(x) is decreasing for all x > a. y

f(a)

f(a+1) f(a+2) f(a+3) y = ¦(x)

x

a+1 a+2 a+3

a

By taking the height of each rectangle to be the value of the function at the left endpoint of each subinterval, we obtain an upper sum U where Z 1 1 P f (x) dx 6 f (i) = U . i=a

a

By taking the height of each rectangle to be the value of the function at the right endpoint of each subinterval, we obtain a lower sum L where Z 1 1 P L= f(i + 1) 6 f (x) dx. i=a

a

Hence, for a function which is decreasing on [a, 1[ , Z 1 1 1 P P f(i + 1) 6 f (x) dx 6 f (i) = U . L= i=a

i=a

a

Similarly, for any continuous function which is increasing on [a, 1[ , Z 1 1 1 P P f(i) 6 f (x) dx 6 f (i + 1) = U . L= i=a

i=a

a

Example 20

Z

1

Write down a series which approximates

2

e¡x dx.

0

Z

1

2

e¡x dx ¼

0

1 P

2

e¡i

i=0

EXERCISE I.2 1 Write down a series which approximates: Z 1 1 a dx p

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e¡x dx

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56

CALCULUS 2

2 Consider the function f(x) = e¡x . a Show that f (x) is decreasing for all x > 0.

Z

1

b Write upper and lower sums that approximate

f(x) dx. 0

c Write an inequality that relates the sums in b to the integral. 3 Consider the function f(x) = ¡

1 . x2

a Show that f (x) is increasing for all x > 0.

Z

1

b Write upper and lower sums that approximate

¡

1

1 dx. x2

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c Write an inequality that relates the sums in b to the integral.

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CALCULUS

J

57

SEQUENCES A sequence is a list of numbers, called terms, in a definite order.

We write the sequence a1 , a2 , a3 , .... as fan g, where an is the nth term of the sequence. The sequence is infinite if it contains an infinite number of terms, so n 2 Z + . The sequence is finite if it contains a finite number of terms, so n 2 Z + , n 6 N . For example, the sequence fan g, where an = f 12 ,

2 3,

3 4,

4 5,

5 6,

....g in this order.

n , n 2 Z + , denotes the infinite set of discrete values n+1

an 1

We can plot an against n to give:

Qw

n

Consider the function f (x) =

x , x+1

x 2 R , x 6= ¡1.

1

x £ x1 lim f (x) = lim x!1 x!1 x + 1 x

= lim

x!1

=

1 1 1+ x

1 1+0

fsince

1 = 0g x!1 x

n =1 n+1

for n 2 Z + .

lim

=1 It follows that

lim

n!1

By considering n sufficiently large, the terms of the sequence will be as close as we like to value 1. We say that L = 1 is the limit of the sequence fan g, where an =

n , n 2 Z +. n+1

This is true even though the terms never actually reach the limit value 1. A sequence fan g has a limit L if for each " > 0 there exists a positive integer N such that jan ¡ Lj < " for all terms an with n > N. We write

lim an = L.

n!1

If the limit of a sequence exists, then we say the sequence converges. Otherwise, the sequence diverges.

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If a sequence converges, then its limit is unique.

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58

CALCULUS

Proof: Suppose fan g is a convergent sequence with two limits L1 , L2 where L1 6= L2 . If " = jL1 ¡ L2 j, then " > 0 since L1 6= L2 . Since Since

" 2

lim an = L1 , there exists N1 2 Z + such that n > N1 ensures jan ¡ L1 j < .

n!1

" 2

lim an = L2 , there exists N2 2 Z + such that n > N2 ensures jan ¡ L2 j < .

n!1

Suppose n = max(N1 , N2 ) is the maximum of N1 and N2 . ) jL1 ¡ L2 j = jL1 ¡ an + an ¡ L2 j 6 jL1 ¡ an j + jan ¡ L2 j

fTriangle Inequalityg

6 jan ¡ L1 j + jan ¡ L2 j
0, there exists N 2 Z + such that N" > 1. lim c = c.

Result 1:

For any real constant c,

Proof:

Let an = c for n 2 Z + .

n!1

) jan ¡ cj = jc ¡ cj = 0 < " for all " > 0. For any " > 0 and for all n > 0 we have jan ¡ cj < ". ) by the definition of a limit of a sequence fan g, lim c = c. n!1

1 = 0. n!1 n

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Result 2:

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CALCULUS

1 for n 2 Z + . n

Let an =

Proof:

59

¯ ¯ ¯1 ¯ 1 ) jan ¡ 0j = ¯ ¡ 0¯ = . n

n

For any " > 0 there exists N 2 Z + such that N " > 1

fArchimedean propertyg

1 < ". N 1 1 < . n N

) For n > N ,

) for any " > 0, there exists N 2 Z + such that for n > N, jan ¡ 0j =

1 1 < < ". n N

) by the definition of a limit of a sequence,

Result 3:

If p > 0, then

Proof:

Let an =

lim

n!1

1 np

lim an = lim

n!1

n!1

1 = 0. n

1 = 0. np

for n 2 Z + . 1

Suppose " > 0 is given. Then " p > 0 and by the Archimedean property there exists 1

N 2 Z + such that N " p > 1. )

1 1 < "p . N

) since y = xp , p > 0, is an increasing function,

1 1 p p < (" ) = ". Np

) for any " > 0, there exists N 2 Z + such that for n > N, ¯ ¯ 1 1 ¯1 ¯ jan ¡ 0j = ¯ p ¡ 0¯ = p 0.

Consider the sequence fcn g, c 2 R .

Result 4:

If 0 6 jcj < 1, the sequence fcn g converges to 0, so

lim cn = 0.

n!1

If jcj > 1, the sequence fcn g diverges. As n ! 1, jcjn ! 1. If c = 0 then

Proof:

lim cn = 0.

fResult 1g

n!1

Let an = cn for n 2 Z + , where c is a constant such that 0 < jcj < 1. If we let jcj =

1 , 1+d

then d =

1 ¡ 1 > 0. jcj

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By the Bernoulli Inequality (see Exercise A question 9), since d > 0, (1 + d)n > 1 + nd > 0 for all n 2 Z + .

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60

CALCULUS

) jcjn =

1 1 1 6 < (1 + d)n 1 + nd nd

for all n 2 Z + .

Given " > 0, "d > 0, so by the Archimedean property there exists N 2 Z + such that N"d > 1. )

1 < ". Nd

) jan ¡ 0j = jcn ¡ 0j = jcn j = jcjn
1 + n" by the Bernoulli Inequality. Now f1 + n"g diverges to infinity as n ! 1, so by comparison, jcjn ! 1.

THE SQUEEZE THEOREM FOR SEQUENCES Suppose we have sequences of real numbers fan g, fbn g, and fcn g where an 6 bn 6 cn for all n 2 Z + . If lim an = lim cn = L exists, then lim bn = L. n!1

n!1

n!1

Suppose L = lim an = lim cn .

Proof:

n!1

n!1

Given " > 0 there exists a natural number N such that if n > N then jan ¡ Lj < " and jcn ¡ Lj < " for all n > N ) ¡ " < an ¡ L < " and ¡" < cn ¡ L < " for all n > N. Now an 6 bn 6 cn , so an ¡ L 6 bn ¡ L 6 cn ¡ L. ) ¡" < bn ¡ L < " for all n > N ) jbn ¡ Lj < " for all n > N. Hence

lim bn = L.

n!1

The Squeeze Theorem still holds even if the condition an 6 bn 6 cn only applies for every natural number from some point n > k. The finite number of sequence terms from n = 1 to n = k does not affect the ultimate convergence (or divergence) of the sequence.

BOUNDED SEQUENCES A sequence of real numbers fan g is said to be bounded if there exists a real number M > 0 such that jan j 6 M for all n 2 Z + . We can deduce that:

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Every convergent sequence is bounded.

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CALCULUS

Let fan g be a sequence where

Proof:

61

lim an = a.

n!1

If we let " = 1, then by the definition of convergence there exists a natural number N such that jan ¡ aj < 1 for all n > N. But from Corollary 3 of the Triangle Inequality, jan j ¡ jaj 6 jan ¡ aj < 1 for all n > N . ) jan j 6 1 + jaj for all n > N. If we define M as the maximum value in the set f1+jaj, ja1 j, ...., jaN ¡1 jg then jan j 6 M for all n 2 Z + . ) the sequence fan g is bounded.

COMPARISON TEST FOR SEQUENCES Consider sequences fan g, fbn g such that 0 6 an 6 bn . If fan g diverges, then fbn g diverges. If fbn g converges, then fan g converges.

ALGEBRA OF LIMITS THEOREMS Suppose fan g converges to a real number a and fbn g converges to a real number b. 1

lim (an + bn ) = lim an + lim bn = a + b

n!1

n!1

n!1

2 The sequence fan bn g converges and

µ 3 If b 6= 0 then

lim

n!1

an bn

¶ =

lim (an bn ) =

³

n!1

lim an

n!1

lim bn

=

n!1

lim an

n!1

´³

´ lim bn = ab.

n!1

a . b

These results can be extended to finite sums and products of limits using mathematical induction. Proof of 2: For n 2 Z + , we have an bn ¡ ab = an bn ¡ an b + an b ¡ ab = an (bn ¡ b) + b(an ¡ a). By the Triangle Inequality, jan bn ¡ abj 6 jan (bn ¡ b)j + jb(an ¡ a)j = jan j jbn ¡ bj + jbj jan ¡ aj Since fan g and fbn g are convergent sequences, they are bounded and there exists M1 , M2 > 0 such that jan j 6 M1 and jbn j 6 M2 for all n 2 Z + . If we let M be the greater of M1 and M2 , then jan bn ¡ abj 6 M jbn ¡ bj + M jan ¡ aj for all n 2 Z + . lim an = a and

Since

n!1

such that jan ¡ aj
N1 , and jbn ¡ bj
N2 .

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62

CALCULUS

We have applied the formal definition of the limit of a sequence to rigorously establish some key results for sequences that can now be used to deal very efficiently with more general problems. Example 21

³ ń 4 5

Suppose an =

+

3 ¡9 n

for all n 2 Z + . Find

lim an .

n!1

By the generalised version of 1 of the Algebra of Limits Theorems, h³ ń i ³ ń 4 3 4 3 lim + ¡ 9 = lim + lim + lim (¡9) 5

n!1

5

n!1

n

n!1

n

n!1

provided each of these limits exist. ¡ ¢n Now lim 45 = 0 fsince 0 < 45 < 1g n!1 ³ ´ 3 1 = lim 3 £ lim =3£0 =0 lim n!1

n!1

n

n!1

n

lim (¡9) = ¡9

n!1

)

h³ ń 4 5

lim

n!1

+

i

3 ¡ 9 = 0 + 0 ¡ 9 = ¡9 n

Example 22 Let an =

2n2 + 4n ¡ 3 n2 ¡ 4 ln n

for all n 2 Z + . Find

lim an .

n!1

4

3

2+ ¡ 2 2n2 + 4n ¡ 3 n n Dividing the numerator and denominator by n , = 4 ln n n2 ¡ 4 ln n 1¡ 2 n ³ ´ 4 3 lim 2 + ¡ 2 n!1 n n ³ ´ ) lim an = 4 ln n n!1 lim 1 ¡ 2 2

n!1

)

lim

n!1

1 =0 n2

lim

Using the limit laws,

n

n!1

³ ´ ³ ´ ³ ´ 4 3 1 1 2 + ¡ 2 = lim (2) + 4 lim ¡ 3 lim 2 n

n!1

n

n!1

n!1

n

n


You can use your calculator to check your answer.

=2+0+0 =2 Now 0 < ln n < n for all n > 1 ln n 1 < n2 n 4 ln n 4 0< 2 < n n

0
an

or an+1 6 an

for all n.

To show that a sequence is monotone we show that either an+1 ¡ an > 0 or that an+1 ¡ an 6 0 for all n 2 Z + . Alternatively, suppose an can be represented by a differentiable function f (x), x 2 R , x > 1 such that an = f(n) for all n 2 Z + . If f 0 (x) > 0 for all x > 1, or if f 0 (x) 6 0 for all x > 1, then fan g is monotone. THE MONOTONE CONVERGENCE THEOREM A monotone sequence of real numbers is convergent if and only if it is bounded. For example: ² The sequence fan g where an = n, n 2 Z + , is a monotone increasing sequence which is unbounded. Clearly fan g diverges. 1

² The sequence fan g where an = 1 ¡ , n 2 Z + , is a monotone increasing sequence which is n ¯ ¯ 1¯ ¯ bounded since jan j = ¯1 ¡ ¯ < 1 for all n 2 Z + . n

We have shown previously that fan g is convergent and

lim an = 1.

n!1

EXERCISE J.2 a Prove that the sequence fun g with nth term un =

1

2n ¡ 7 3n + 2

is:

i monotone increasing ii bounded. b Determine whether the following sequences are monotone and bounded, and calculate their limits if they exist: n n o n o 3 1 i ii n n ¡n

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65

CALCULUS

½ 2 Prove that the sequence

¾

1 £ 3 £ 5 £ :::: £ (2n ¡ 1) 2n n!

3 The sequence fxn g is defined by x1 = 0, xn =

is convergent.

p

4 + 3xn¡1 .

a Use mathematical induction to show that fxn g is monotone increasing. b Evaluate x1 , x2 , x3 , ...., x11 , and hence suggest an upper bound for fxn g. c Use mathematical induction to prove that fxn g is bounded. d Hence find

lim xn .

n!1

a Find the values of 1 + 11 , 1 +

4

1 1 , 1+ 1 1 + 11 1+

, 1+

1 + 11

1 1

1+

1+

1 1+

1 1

b Give a recursive definition for the sequence above in terms of un . c Show that fun g is bounded but not monotone. d Given that fun g converges, find the exact value of

lim un .

n!1

a Consider the sequence fun g defined by u1 = a (a positive constant) and ³ ´ 2 for all n 2 Z + . un+1 = 12 un +

5

un

Use your calculator to find u1 , u2 , u3 , ...., u8 when: i a=5 ii a = 1

iii a of your choice.

b Is fun g monotone for all a > 0? c Do your experimental results in a suggest the existence of a limit L such that If so, find L. d If u1 = a where a > 0 and un+1 =

2!

3!

n

n

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ii Show that en < 1 + 1 + + + :::: + < 1 + 1 + + 2 + :::: + n¡1 for all 2! 3! n! 2 2 2 + n2Z . iii Hence show that fen g is convergent. ³ ´ ³ ´ 1 n 1 n i Given that lim 1 + = e ¼ 2:718, show that lim 1 ¡ = e¡1 . n!1 n!1 n n ³ ´ n! ii Hence show that lim = 0 using the Squeeze Theorem. n

5

d

lim un

n!1

for all n 2 Z + .

i Show that 2 6 en < en+1

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n

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6

un

75

f

³ ´ 3 un + , find

given that it exists. ³ ´ k State lim un given that fun g is defined by u1 = 6 and un+1 = 12 un + . n!1 un p Suggest a recurrence relationship for generating values of 3 k. ³ ´ 1 n Expand 1+ , n 2 Z + , using the Binomial Theorem. n ³ ´ 1 n Define fen g by en = 1 + and show that en equals: n ³ ´ ³ ´³ ´ h³ ´³ ´ ³ í 1 1 1 1 2 1 1 2 n¡1 1¡ + 1¡ 1¡ + :::: + 1¡ 1¡ :::: 1 ¡ 1+1+

25

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lim un = L.

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66

CALCULUS

K

INFINITE SERIES

Let fu1 , u2 , u3 , ....g be an infinite sequence. We can form a new sequence fSn g = fS1 , S2 , S3 , ....g by letting S1 = u1 S2 = u1 + u2 .. .

n P

Sn = u1 + u2 + :::: + un =

ui .

i=1

The value Sn , which is the sum of the first n terms of fun g, is called the nth partial sum. Each term of fSn g is a finite series. If

1 P

lim Sn =

n!1

un

exists and equals some finite value S, then the infinite series is convergent.

n=1

Otherwise it is divergent. Example 24 Let fun g be defined by un = rn¡1 where r 2 R , r 6= 0, n 2 Z + . Find an expression for Sn , the nth partial sum of fun g, which does not involve a summation. Sn =

n P

n P

ui =

i=1

ri¡1 = 1 + r + r2 + :::: + rn¡1

i=1

rSn = r + r2 + r3 + :::: + rn

) )

rSn ¡ Sn = rn ¡ 1 )

rn ¡ 1 r¡1

Sn =

It is often important to know when

lim Sn =

n!1

1 P

un

exists, and if so, what its value is. In general

n=1

it is not possible to write Sn as an explicit expression as we did in Example 24. However, we shall see that more difficult functions can often be expressed as simpler infinite series. Great mathematicians such as Euler and Newton did much of their foundation work using infinite series representations of functions, though it was not until much later that other mathematicians such as Cauchy and Lagrange rigorously established when such representations were valid. Since convergence of a series is in effect convergence of a sequence of partial sums, many of the sequence results apply. For example: an

and

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bn

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(an § bn ) =

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n=1

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1 P

1 P

0

can = c

n=1

²

are convergent series, then

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1 P

50

²

bn

n=1

5

n=1

1 P

75

1 P

If

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CALCULUS

67

However, because the form of the sequence of partial sums is generally too unwieldy to deal with using our earlier methods, we need a special set of tests and conditions for determining when the limits of these partial sums exist. We start with a useful result which can tell us something either about a series

1 P

an or its associated

n=1

sequence of general terms fan g: 1 P

If the series

an

is convergent then

n=1

lim an = 0.

n!1

Let Sn = a1 + a2 + :::: + an ) an = Sn ¡ Sn¡1

Proof:

Now

1 P

an is convergent, so fSn g is convergent (by definition).

n=1

Letting

lim Sn = S,

lim Sn¡1 = S

n!1

n!1

lim an = lim (Sn ¡ Sn¡1 ) = S ¡ S = 0

)

n!1

n!1

1 = 0, n!1 n

lim

We shall show later that even though

1 P 1 n=1

n

diverges extremely slowly.

Therefore, the converse of the above theorem is not true. However, we may establish the following Test for Divergence: If

lim an

n!1

lim an 6= 0, then the series

does not exist, or if

n!1

1 P

an

is divergent.

n=1

Example 25 1 P

n2 2+4 5n n=1

Show that the series

The nth term of the series is an = n2 n!1 5n2 + 4 1

diverges. n2 . 5n2 + 4

lim an = lim

)

n!1

= lim

n!1

= Since

5 + n42

1 5

lim an 6= 0, the series diverges.

n!1

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The Test for Divergence puts no sign restriction on each term of fan g. However, all of the following series tests only apply to series with positive terms.

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68

CALCULUS

THE COMPARISON TEST FOR SERIES Let fan g be a positive sequence, so an > 0 for all n. 1 1 P P bn such that an 6 bn , then an is also convergent. If there exists a convergent series n=1

1 P

Conversely, if an > bn and

n=1

1 P

bn diverges, then so does

n=1

an .

n=1

Proof of the first part: Let fAn g and fBn g be the sequences of partial sums associated with an and bn respectively. Since an , bn > 0, fAn g and fBn g are monotonic increasing. Suppose an 6 bn , and that fBn g is convergent with

lim Bn = B.

n!1

) 0 6 An 6 Bn 6 B. ) An is also a bounded monotonic sequence, and therefore converges by the Monotone Convergence Theorem. With a minor adjustment to the proof, the result can be shown to hold if an > 0 for all n. 1 P However, the difficulty in using the Comparison Test is in finding a suitable bn . n=1

An appropriate geometric series often tends to work. Indeed, convergent geometric series are used in the proofs of some of the most general and important convergence tests. GEOMETRIC SERIES 1 P A series ak is a geometric series if there exists a constant r, called the common ratio of the series, k=1

such that ak+1 = rak for all k 2 Z + , and a1 = a is a constant. 1 1 P P In this case ak = a + ar + ar2 + ar3 + :::: = ark¡1 . k=1

k=1

n P

= a(1 + r + r2 + :::: + rn¡1 ). 1 1 P P rk¡1 , or equivalently rk . Since a 2 R is a constant it suffices to examine

The nth partial sum Sn =

ar

k¡1

k=1

k=1

If jrj < 1, then the geometric series

1 P

k=0

rk converges with sum S =

k=0

1 P k=0

rk =

1 . 1¡r

If jrj > 1 then the geometric series diverges. Proof: 1 P

If r = 1 then

1k¡1 = 1 + 1 + 1 + 1 + ...., and the nth partial sum is

k=1

Sn =

n P

1k¡1 = 1 + 1 + :::: + 1 = n. | {z } k=1 n times

1 P

Now lim Sn = lim n DNE, so

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1k diverges.

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69

CALCULUS

If r = ¡1 then

½

1 P

k¡1

(¡1)

= 1 ¡ 1 + 1 ¡ 1 :::: and the nth partial sum is Sn =

k=1

1, n odd 0, n even.

DNE as Sn alternates between constants 1 and 0 as n ! 1.

lim Sn

n!1

n P

If jrj 6= 1 then Sn =

rk¡1 = 1 + r + :::: + rn¡1

rn ¡ 1 r¡1 rn ¡ 1 lim Sn = lim n!1 n!1 r ¡ 1 k=1

)

=

ffrom Example 24g

lim rn = 0 for jrj < 1

By Result 4 of the limit theorems for sequences,

n!1

lim rn DNE for jrj > 1.

and Hence for jrj < 1 For jrj > 1,

lim Sn =

n!1

lim Sn

1 , 1¡r

1 P

and so

rk =

k=0

1 P

DNE and so

n!1

n!1

rk

1 1¡r

is convergent.

diverges.

k=0

Example 26 Test the series

1 P

1 n+1 2 n=1

for convergence.

Now 2n is positive for all n, and 2n + 1 > 2n . ¡ ¢n 1 1 < n = 12 for all n 2 Z + . ) 0< n 2 +1

But

2

1 ¡ ¢n P 1 n=1

is a convergent geometric series and therefore,

2

1 P

1 n+1 2 n=1

by the Comparison Test,

also converges.

We cannot use the Comparison Test in the same way as in Example 26 to test the series

1 P n=1

2n

1 ¡1

for convergence. However, the following test may be useful when the Comparison Test cannot be applied directly.

THE LIMIT COMPARISON TEST Suppose that

1 P

an

1 P

and

n=1

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1 P

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3 If

an = c, c > 0, then the series either both converge or both diverge. bn 1 1 P P a lim n = 0 and bn converges, then an converges. n!1 bn n=1 n=1

lim

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2 If

are series with positive terms.

n!1

100

1 If

bn

n=1

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70

CALCULUS

Proof of 1: c 2

Since c > 0, 0 < " = . Since

an = c, bn

lim

n!1

using the definition of a limit there exists N 2 Z + such that

¯ ¯ ¯ an ¯ c ¡c¯ < ¯

)

)

for all n > N

bn 2 c an c ¡ < ¡c< 2 bn 2 c a 3c ) < n < 2 b n ³ ´ ³2 ´ c 3c bn < an < bn 2 2 1 P

Now if

bn

for all n > N

n=1

an

bn

1 ³ ´ P c

diverges then so does

n=1

n=1

1 P

Hence by the Comparison Test,

bn .

also converges.

n=1

1 P

2

n=1

1 P

Hence by the Comparison Test, However, if

1 ³ ´ P 3c

converges then so does

an

2

bn .

also diverges.

n=1

Example 27 Test the series

Let an = )

1 P

1 n¡1 2 n=1

1 2n ¡ 1

for convergence or divergence.

and bn =

1 . 2n

an 2n = lim n n!1 bn n!1 2 ¡ 1 1

lim

= lim

n!1

=1 Since

1 P

1 n 2 n=1

1¡

³ ń 1

n 2 since

lim

¡ 1 ¢n

n!1

2

o =0

is a convergent geometric series,

1 P

1 n¡1 2 n=1

fLimit Comparison Test 1g

converges also.

THE INTEGRAL TEST The Integral Test links the sum of a series to the integral of a positive function. Z 1 1 P We have seen that if a is an integer, f (i) ¼ f (x) dx i=a

f (i) ¼

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f (x) dx

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In particular, when a = 1,

a

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1 P

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CALCULUS

71

The Integral Test is: Suppose that f is a continuous, positive, decreasing function on [ 1, 1 [ . Let an = f(n), n 2 Z + , so that the terms in fan g are all positive. Z 1 1 P f (x) dx is convergent, then an is convergent. 1 If 1

Z

n=1

1

2 If

f (x) dx is divergent, then

1

1 P

an

is divergent.

n=1

Z

1

Clearly this test is only of practical use if

f (x) dx can be evaluated relatively easily.

1

Proof of 1: If f(x) is a continuous, positive, decreasing function, then we can approximate the integral Z 1 f (x) dx using lower and upper sums: 1

a1 a2

a2

y = f(x)

a3

y = f(x) an-1

an 1

2

3

n

The lower sum

Z

a2 + a3 + :::: + an + :::: 6

1

n=1

1

Hence, 1

Z

1

If

f (x) dx 6

1 P

2

3

n

The upper sum

Z 1 f (x) dx a1 + a2 + :::: + an¡1 + :::: > 1 Z 1 1 P ) f (x) dx 6 an

1

f (x) dx 1 Z 1 1 P an 6 a1 + f (x) dx

)

Z

1

Z

an 6 a1 +

1

1

n=1

f (x) dx

1

n=1

f (x) dx converges then the sequence of partial sums fSn g of

1

1 P

an

is bounded and

n=1

monotonic increasing, and hence convergent also. From the proof we also gain the useful result: If f(x) is a continuous, positive, decreasing function on [1, 1 [ , then Z 1 Z 1 1 P f(x) dx 6 an 6 a1 + f (x) dx.

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72

CALCULUS

Example 28 1 P

1 2+1 n n=1

Test

f (x) =

1 x2 + 1

for convergence.

is continuous, positive, and decreasing for x > 1.

) the conditions to use the Integral Test are satisfied. Z 1 Z b 1 1 dx = lim dx Now 2 2 x +1

1

b!1


x +1

1

= lim [arctan x]b1 b!1 ¡ ¢ = lim arctan(b) ¡ ¼4 b!1 ¼ ¼ 2 ¡ 4

=

Z )

1

=

You can use your calculator to estimate

Z

¼ 4

1 P

f (x) dx is convergent, and so is

1

n=1

1

f (x) dx

1

1 . n2 + 1

Example 29 1 P 1

For what values of p is the series

n=1

If p < 0 then

lim

n!1

If p > 0 then

1 = lim njpj n!1 np

lim

In each of these cases,

n!1

which diverges, and if p = 0 then

1 6= 0, np

1 = 0. np

lim

n!1

convergent?

np

lim

n!1

1 = 1. np

so by the Test for Divergence, the series diverges.

Since the function f(x) =

1 xp

is continuous, positive, and

decreasing for x > 1, we can apply the Integral Test: Z 1 Z b 1 1 dx = lim dx p p 1

b!1

x

·

1

x

¸b

1 x1¡p for p 6= 1 = lim b!1 1 ¡ p 1 1 1 lim b1¡p ¡ = 1 ¡ p b!1 1¡p

(

=

¡

1 1¡p

if p > 1

if 0 < p < 1 Z b 1 1 dx = lim dx = lim [ln jxj]b1

DNE

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= lim (ln b) which DNE b!1

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1 P

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) by the Integral Test, the series

b!1

x

1

5

b!1

x

1

50

1

75

Z For p = 1,

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73

CALCULUS

p-SERIES

1 P 1

The series

n=1

this form.

is called the p-series, and can be used to rapidly test the convergence of series of

np

As shown in Example 29: 1 P 1

The p-series


np

n=1

1 P 1

1 P 1 p = 0:5 n n n=1 n=1

For example, the series

1 P 1

The p-series with p = 1 is

n=1

n

1 2

=1+

is divergent since p = 1 3

+

1 4

+

1 P

Test for convergence or divergence

sin

lim an = lim

)

lim

1 . n sin

n!1

1 = 0. n

n

³ ´ 1 n

1 n

n!1

n!1

³ ´ 1 .

n=1

Let x =

< 1.

+ :::: and is called the harmonic series.

This is an example of a series which is divergent even though Example 30

1 2

= lim

x!0

sin(x) =1 x

by the Fundamental Trigonometric Limit.

1 P

) by the Limit Comparison Test

sin

³ ´ 1 n

n=1

diverge.

1 P 1

Since the harmonic series

n=1 n

n=1

1 P

diverges,

1 P 1

and sin

n

either both converge or both

³ ´

n=1

1 n

also diverges.

EXERCISE K.1 1 Determine whether the following series converge or diverge using an appropriate geometric series, the Comparison Test, or the Test for Divergence. ´ 1 1 1 1 ³ P P P P 1 n2 3n + 2n 1 1 a b c d ¡ 2n n 2 n=1

e

n=1

3(n + 1)(n + 2)

n=1

2

2 Use the Limit Comparison Test with bn = p 3 Determine whether

1 n n n=1

1 P

1 n! n=1

and

n=1

to show that the series

n3

1 P

6

n

n

1 P 2n2 + 3n is convergent. p n=1

5 + n7

are convergent using the Comparison Test.

4 Determine whether the following series converge or diverge using the Comparison Test or Limit Comparison Test. 1 1 1 P P P 1 1 sin2 n a b c p p p 3

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n(n + 1)(n ¡ 1)

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n(n + 1)(n + 2) p 1 P n d n ¡ 1 n=2 n=1

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n n

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1 ln n n=2



74

CALCULUS 1 P

5 Find all the values of x 2 [0, 2¼] for which the series

2n jsinn xj converges.

n=0

1 P

6 Find c if

(1 + c)¡n = 2.

n=2

7 Use the Integral Test to determine whether the following series converge: 1 1 1 P P P n ln n ¡n2 a b ne c 2 n=1 n + 1 ¼ 4

8 Show that

n=1 n

n=1

p + p + :::: + p = p . c Hence prove that the sequence fSn g of partial sums diverges. 1 P 1 d Does converge or diverge? Explain your answer. p n

n=1

12 Consider the series

1 P n2 n=1

3n

. n P i2

a Use induction to prove that the nth partial sum is Sn =

i=1

b Find

3i

=

3 ¡ 3¡n (n2 + 3n + 3) . 2

lim Sn .

n!1

c Does the series series.

1 P n2 n=1

3n

converge? If not, explain why. If so, find the value of the sum of the

APPROXIMATING BY TRUNCATING AN INFINITE SERIES The tests for infinite series can help us determine whether a series is convergent or divergent. However, even if we prove an infinite series is convergent, we do not in general have techniques to 1 P determine the value of the limit S = an . It is only in special cases such as geometric series that n=1

we can do this.

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Instead, we can approximate the sum S of a convergent infinite series using a partial sum Sk .

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CALCULUS

Suppose we approximate the sum of a convergent series S = 1 P

an ¼

n=1

k P

1 P

75

an by the sum Sk of its first k terms:

n=1

for some k 2 Z + .

an = a1 + a2 + :::: + ak

n=1

If f is a continuous, positive, decreasing function on [ k, 1 [ , then we can apply the Integral Test. 1 P an and the error satisfies The error in the approximation is Rk = S ¡ Sk = n=k+1 Z 1 Z 1 f(x) dx < Rk < f (x) dx. k+1

k

Example 31 1 P

Suppose we can use the Integral Test to show that

an

is convergent, where an = f (n).

n=1

1 P a Show that the error Rk in approximating an n=1 Z 1 Z 1 satisfies f(x) dx < Rk < f (x) dx. k+1

by a1 +a2 +::::+ak

k

1 P 1

b Hence determine the number of terms necessary to approximate decimal places. a The error Rk = S ¡ Sk =

1 P

k P

an ¡

n=1

n=1

n3

for some k 2 Z +

correct to two

y

an

a1

n=1

= ak+1 + ak+2 + ak+3 + ::::

a2

y = f(x)

a3

From the areas of lower rectangles, we deduce Z 1 f (x) dx. Rk = ak+1 + ak+2 + ak+3 + ::::
f(x) dx

Z

1

Hence

Z f (x) dx < Rk
10 fas k > 0g

n3

to 2 decimal places.


Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\075IB_HL_OPT-Calculus_01.cdr Tuesday, 26 February 2013 12:40:51 PM BRIAN

76

CALCULUS

EXERCISE K.2 1

1 P

1 2 5n n=1

a Estimate the error when

is approximated by its first 12 terms. 1 P 1

b How many terms are necessary to approximate

n=1

1 P

2 The nth partial sum of a series

is Sn =

an

n=1

a Find an .

b Write

1 P

correct to 6 decimal places?

n4

n¡1 . n+1

an

in expanded form and find its sum.

n=1

3

a Use your calculator to evaluate the partial sums S1 , S2 , S3 , S4 , 1 P n S5 , and S6 for . Give your answers in rational form. n=1

(n + 1)!

b Conjecture a formula for Sn . c Use mathematical induction to prove your conjecture. 1 P n d Hence find . n=1


(n + 1)!

1 P 1

4 The harmonic series is defined by

n=1 n

=1+

1 2

+

1 3

+

1 4

+ :::: .

Consider the following sequence of partial sums for the harmonic series: S1 = 1 In Example 29 we saw the S2 = 1 + 12 most efficient proof that the ¡ ¢ Harmonic series is divergent. S4 = 1 + 12 + 13 + 14 ¡1 1¢ This is an alternative proof. 1 2 >1+ 2 + 4 + 4 =1+ 2 ¡ ¢ ¡ ¢ S8 = 1 + 12 + 13 + 14 + 15 + 16 + 17 + 18 ¡ ¢ ¡ ¢ > 1 + 12 + 14 + 14 + 18 + 18 + 18 + 18 = 1 + 32 a Use the same method to find an inequality involving S16 . b Conjecture an inequality involving S2m , m 2 Z + . Use mathematical induction to prove your conjecture. c Show that S2m ! 1 as m ! 1 and hence prove that fSn g is divergent.

ALTERNATING SERIES Thus far, we have dealt with series with only positive terms. An alternating series is one whose terms are alternately positive and negative. For example, 1 ¡

1 2

+

1 4

¡

1 8

+

1 16

¡

1 32

+ ::::

THE ALTERNATING SERIES TEST If the alternating series

1 P

(¡1)n¡1 bn = b1 ¡ b2 + b3 ¡ :::: satisfies 0 6 bn+1 6 bn

n=1

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for all n 2 Z + , and if

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CALCULUS

77

The theorem also applies if the first term is negative, since we can multiply through by constant ¡1 and proceed as above. Proof: The (2n + 2)th partial sum of the series is S2n+2 = b1 ¡ b2 + :::: ¡ b2n + b2n+1 ¡ b2n+2 , where the bi are all non-negative and non-increasing. We therefore find that S2n+1 = S2n + b2n+1 S2n+2 = S2n + b2n+1 ¡ b2n+2 S2n+3 = S2n+1 ¡ b2n+2 + b2n+3 = S2n+2 + b2n+3 Since b2n+1 > b2n+2 > b2n+3 , we find S2n+1 > S2n+3 > S2n+2 > S2n . Also, S2n+2 = (b1 ¡ b2 ) + (b3 ¡ b4 ) + :::: + (b2n+1 ¡ b2n+2 ). Because the bi are non-increasing, each expression in brackets is > 0. Hence Sn > 0 for any even n, and since S2n+1 > S2n+2 , Sn > 0 for all n. Finally, since S2n+1 6 b1 , we conclude that b1 > S2n+1 > S2n+3 > S2n+2 > S2n > 0. Hence the even partial sums S2n and the odd partial sums S2n+1 are bounded. The S2n are monotonically non-decreasing, while the odd sums S2n+1 are monotonically non-increasing. Thus the even and odd series both converge. Since S2n+1 ¡ S2n = b2n+1 , the sums converge to the same limit if and only if

lim bn = 0.

n!1

The convergence process is illustrated in the following diagram. 0

b1

0

b1

b 1- b 2

0

b 1- b 2 + b3

0 b 1- b 2 + b3 - b4

b1 b1

even partial sums odd partial sums 0

b1

lim Sn

n®1

If 0 6 bn+1 6 bn

for all n 2 Z + but

between the two values

lim S2n

n!1

lim bn 6= 0, then the series will eventually oscillate

n!1

lim S2n+1 .

and

n!1

even partial sums

odd partial sums

0

b1 lim S2 n

lim S2 n + 1

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78

CALCULUS

Example 32 Show that 1 ¡ 12 + 13 ¡ 14 + :::: =

1 P (¡1)n¡1

1 P

This is an alternating series of the form where bn =

Compare this with the harmonic series which diverges!

(¡1)n¡1 bn ,

n=1

1 . n

1 1 < , n+1 n

Since

converges.

n

n=1

the series satisfies 0 < bn+1 < bn

for all n 2 Z + . lim bn = lim

Also,

n!1

n!1

1 P (¡1)n¡1

)

n

n=1

1 =0 n

converges by the Alternating Series Test.

Suppose a convergent infinite series converges to a sum S. If the nth partial sum Sn is used to estimate the sum S, the truncation error is defined by Rn = jS ¡ Sn j. THE ALTERNATING SERIES ESTIMATION THEOREM Suppose S =

1 P

(¡1)n¡1 bn

is the sum of a convergent alternating series,

n=1

for all n 2 Z + and

so 0 6 bn+1 6 bn

lim bn = 0.

n!1

The truncation error Rn = jS ¡ Sn j 6 bn+1 . Proof: S ¡ Sn =

1 P

(¡1)k¡1 bk ¡

k=1

n P

(¡1)k¡1 bk

k=1 n+1

= (¡1)n bn+1 + (¡1)

bn+2 + ::::

= (¡1) [(bn+1 ¡ bn+2 ) + (bn+3 ¡ bn+4 ) + ::::] n

Since br > br+1

for all r 2 Z + , bn+r > bn+r+1

for all r 2 Z + .

) (bn+1 ¡ bn+2 ) + (bn+3 ¡ bn+4 ) + :::: > 0 ) Rn = jS ¡ Sn j = (bn+1 ¡ bn+2 ) + (bn+3 ¡ bn+4 ) + :::: = bn+1 ¡ (bn+2 ¡ bn+3 ) ¡ (bn+4 ¡ bn+5 ) ¡ :::: = bn+1 ¡ [(bn+2 ¡ bn+3 ) + (bn+4 ¡ bn+5 ) + ::::]

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[(bn+2 ¡ bn+3 ) + (bn+4 ¡ bn+5 ) + ::::] > 0

since

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CALCULUS

79

Example 33 1 P (¡1)n¡1

Find the sum of

correct to 3 decimal places.

n!

n=1

1 P

This is an alternating series of the form

(¡1)n¡1 bn , where bn =

n=1

1 1 0< < , (n + 1)! n!

Now

Also, 0 < ) since

1 1 < n! n 1 lim = 0, n!1 n

so 0 < bn+1 < bn

lim

n!1

1 . n!

for all n 2 Z + .

1 = lim bn = 0 n!1 n!

fSqueeze Theoremg

) the series converges by the Alternating Series Test. Now S = 1 ¡

1 2

+

Notice that b7 =

1 6

¡

1 5040

and S6 = 1 ¡

1 24

+

1 120

¡

1 720

1 2000 = 0:0005 1 1 1 1 2 + 6 ¡ 24 + 120

+

1 5040

¡

1 720

+ ::::

0 for all n, absolute convergence is the same as convergence. 1 P (¡1)n¡1

A series such as

n

n=1

which is convergent but not absolutely convergent, is called

conditionally convergent. So what is important about absolute and conditional convergence? We all know that the addition of scalars is commutative, so a + b = b + a. Furthermore, if we have a N P an , then we can also reorder the terms without affecting the total sum. Infinite series finite sum n=1

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which are absolutely convergent behave like finite series, in that we can reorder the terms of the series without affecting the sum. However, the same is not true for conditionally convergent series!

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80

CALCULUS

S =1¡

For example, let

1 2S 1 2S 3 2S 3 2S

) ) Adding (1) and (2) gives )

=

1 2

¡

=0+ =1+ =1+

1 2

1 3

+

¡

1 4

+

1 5

¡

1 6

+

1 7

1 8

¡

+ :::: (1)

1 1 1 4 + 6 ¡ 8 + :::: 1 1 1 1 2 + 0 ¡ 4 + 0 + 6 + 0 ¡ 8 + :::: 0 + 13 ¡ 12 + 15 + 0 + 17 ¡ 14 + :::: 1 1 1 1 1 3 ¡ 2 + 5 + 7 ¡ 4 + ::::

(2)

We thus obtain a rearrangement of the original series, but with a different sum! In fact, Riemann showed that by taking groups of sufficiently large numbers of negative or positive terms, it is possible to rearrange a conditionally convergent series so it adds up to any arbitrary real value. THEOREM OF ABSOLUTE CONVERGENCE 1 P

If a series

an

is absolutely convergent then it is convergent.

n=1

By the definition of absolute value, ¡ jan j 6 an 6 jan j

Proof:

) 1 P

Now if

0 6 an + jan j 6 2 jan j

is absolutely convergent then 2

an

n=1

jan j is convergent.

n=1

1 P

) by the Comparison Test,

1 P

(an + jan j) is convergent.

n=1

1 P

But

1 P

an =

n=1

n=1

1 P

) since

(an + jan j) ¡

1 P

jan j since the series is absolutely convergent.

n=1

1 P

(an + jan j) and

n=1

jan j are both convergent,

n=1

1 P

an

is convergent.

n=1

Example 34 1 P cos n

cos 1 cos 2 + 2 + :::: has terms with different signs, but is not an alternating series. 12 2

¯ ¯ 1 ¯ cos n ¯ However, ¯ 2 ¯ 6 2 n

n

for all n 2 R , and

1 P 1 n=1

) by the Comparison Test,

¯ 1 ¯ P ¯ cos n ¯ ¯ 2 ¯ is convergent. n

n=1

) by the Theorem of Absolute Convergence,

1 P cos n

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is convergent.

n2

n2

is also convergent.

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2 n=1 n

50

1 P cos n

is convergent.

n2

n=1

75

Show that

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CALCULUS

81

THE RATIO TEST The Ratio Test can be used to determine whether a general series is absolutely convergent, and hence convergent:

¯ ¯ 1 P ¯a ¯ lim ¯ n+1 ¯ < 1, then an is absolutely convergent. n!1 an n=1 ¯ ¯ 1 P ¯a ¯ 2 If lim ¯ n+1 ¯ > 1, then an is divergent. n!1 an n=1 ¯ ¯ ¯a ¯ 3 If lim ¯ n+1 ¯ = 1, the Ratio Test is inconclusive. 1 If

n!1

an

Proof of 1: Let un = jan j, with an 6= 0 for all n 2 Z + . u

Suppose that lim n+1 = L < 1, so given " > 0 there exists a positive integer N ¯ n!1 ¯un ¯u ¯ such that ¯ n+1 ¡ L¯ < " for all n > N. un

In particular, as L < 1 we can choose r such that L < r < 1 and let " = r ¡ L > 0. ¯ ¯ ¯u ¯ Now ¯ n+1 ¡ L¯ < " )

)

)

un un+1 ¡L 1, we let " = L ¡ 1. n!1 an ¯ ¯ ¯ jan+1 j ¯ + ¯ such that ¯ ¡ L¯¯ < " for all n > N ) there exists N 2 Z If

jan j

jan+1 j ¡L 0. b Show that S4 > S2 . Hence prove that in general, S2n > S2n¡2 and 0 6 S2 6 S4 6 :::: 6 S2n 6 :::: . c Show that S2n = b1 ¡ (b2 ¡ b3 ) ¡ (b4 ¡ b5 ) :::: ¡ (b2n¡2 ¡ b2n ) ¡ b2n and S2n 6 b1 . d Hence prove that S2n is convergent. Let lim S2n = S. n!1

e Show that S2n+1 = S2n + b2n+1 . lim bn = 0 then

f Show that if

lim S2n+1 = S

n!1

n!1

and hence

lim Sn = S.

n!1

7 Determine whether these series are absolutely convergent, conditionally convergent, or divergent: 1 1 1 P P P (¡3)n 2n arctan n a b (¡1)n 2 c (¡1)n 3 n=1

d

n=1

g

n!

1 P

n +1

n=1

´ 1 ³ P 1 ¡ 3n n

1 P (¡1)n¡1 ln n

e

3 + 4n

f

n

n=1

n

n=1

1 P (¡1)n n=2

n ln n

¡p p ¢ (¡1)r¡1 r+1¡ r

r=0

8

1 P xn

a Show that

n=0

b Deduce that

n!

lim

n!1

converges for all x 2 R . xn =0 n!

for all x 2 R .

9 Test these series for convergence or divergence:

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n(n + 1)

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84

CALCULUS 1 P 1

10 Test the series

1 P 1

for absolute convergence using the Ratio Test. ¯ ¯ ¯a ¯ Hence explain why the Ratio Test is inconclusive when lim ¯ n+1 ¯ = 1. n=1

and

n2

n=1

n

n!1

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POWER SERIES 1 P

A power series is a series of the form 1 P

or more generally

cn xn = c0 + c1 x + c2 x2 + ::::

n=0

cn (x ¡ a)n = c0 + c1 (x ¡ a) + c2 (x ¡ a)2 + :::: .

n=0

The convergence of a power series will often, but not always, depend on the value of x. 1 P

For example, consider the power series

where cn = 1 for all n. This is in fact the

cn xn

n=0

geometric series 1 + x + x2 + x3 + x4 + :::: , which converges for all jxj < 1. We use the Ratio Test to determine the convergence of a power series. Example 36 1 P (x ¡ 3)n

For what values of x is

n

n=1

(x ¡ 3)n an = n

If

convergent?

¯ ¯ ¯ ¯ ¯ n ¯ an+1 ¯ ¯¯ (x ¡ 3)n+1 ¯ then ¯ £ ¯=¯ n an n+1 (x ¡ 3) ¯ ¯ ¯ ¯ (x ¡ 3)n ¯ =¯ ¯ n+1 ¯ ¯ ¯ ¯ ¯ (x ¡ 3) ¯ =¯ ¯ ¯ 1+ 1 ¯ n ¯ ¯ ¯a ¯ ) lim ¯ n+1 ¯ = jx ¡ 3j n!1

1 P

By the Ratio Test,

an

is divergent if jx ¡ 3j > 1, but is absolutely convergent and hence

an

n=1

jx ¡ 3j < 1

convergent if

)

¡1 < x ¡ 3 < 1 )

2<x R. The power series

1 P

cn (x ¡ a)n

has radius of convergence R if R is the greatest number such

n=0

that the series converges for all x 2 R such that jx ¡ aj < R and diverges for all x 2 R such that jx ¡ aj > R. If the series converges only when x = a, we say R = 0. If the series converges for all x 2 R , we say R is infinite. The interval of convergence I is the set of all points x for which the power series converges. Most of the interval of convergence may be deduced from the radius of convergence. However, we need to consider convergence for the cases jx ¡ aj = R separately. 1 P (x ¡ 3)n

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For instance, we saw in Example 36, that

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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\085IB_HL_OPT-Calculus_01.cdr Friday, 8 February 2013 12:15:18 PM BRIAN

86

CALCULUS

Example 37 1 P (¡3)n xn

Find the radius and interval of convergence for

. p n+1

n=0

If

(¡3)n xn an = p n+1

¯ p ¯ ¯ ¯ n + 1 ¯¯ ¯ an+1 ¯ ¯¯ (¡3)n+1 xn+1 £ ¯ ¯=¯ p an (¡3)n xn ¯ n+2 r n+1 = 3 jxj

then

r

n+2

= 3 jxj

2 1+ n

¯ ¯ ¯a ¯ lim ¯ n+1 ¯ = 3 jxj

)

n!1

1 P

By the Ratio Test,

1 1+ n

an

1 3

converges if jxj
13 .

) the radius of convergence is R = 13 . To determine the interval of convergence, we must consider what happens when x = § 13 . ³ ń 1 1 1 (¡3)n ¡ 3 1 P P P 1 an = = If x = ¡ 13 , p p n=0

n+1

n=0

1 P

Letting r = n + 1,

an =

n=0

1 P

If x = 13 ,

an =

n=0

1 (¡3)n P

³ ń

which is a divergent p-series.

1 3

1 P (¡1)n n=0

1 0:5 r r=1

p n+1

n=0

=

n+1

n=0

1 P

p n+1

So, the interval of convergence of

which converges by the Alternating Series Test. 1 P

is ] ¡ 13 ,

an

n=0

1 3

].

Example 38 a Find the radius of convergence and interval of convergence for lim

b Hence show that a If

n!1

xn n!

1 P xn n=0

n!

, where x 2 R .

= 0 for all x 2 R .

¯ ¯ ¯ ¯ ¯ xn+1 n! ¯¯ ¯ an+1 ¯ ¯ then lim ¯ £ n¯ ¯ = lim ¯ n!1 n!1 (n + 1)! an x

xn an = n!

= lim

n!1

jxj n+1

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CALCULUS

87

DIFFERENTIATION AND INTEGRATION OF POWER SERIES A power series can be differentiated or integrated term by term over an interval contained entirely within its interval of convergence. In the case of differentiation, an open interval is required. If f (x) =

1 P

n

cn x

f 0 (x)

then

=

n=0

1 P

Z

n¡1

ncn x

f (x) dx =

and

n=1

1 P

cn xn+1 . n +1 n=0

Example 39 ¶ Z 0:1 µ 1 P (¡3)n xn p dx. Find 0

n+1

n=0

1 P (¡3)n xn

has interval of convergence ] ¡ 13 ,

p n+1 n=0

From Example 37, the series

1 3

].

) since [0, 0:1] lies entirely within the interval of convergence, ¶ µZ 0:1 ¶ Z 0:1 µ 1 1 P (¡3)n xn P (¡3)n xn p dx = p dx 0

n+1

n=0

0

n=0

·

1 P (¡3)n

=

n=0

p n+1

n+1

¸ 0:1

xn+1 n+1 0

1 P (¡3)n (0:1)n+1

=

n=0

(n + 1)

3 2

EXERCISE K.4 1 Use the geometric series to write a formula for each of the following. In each case find the corresponding radius of convergence and interval of convergence. 1 1 1 P P P a x3n b (2 ¡ x)n c (¡1)n x4n n=0

n=0

n=0

2 Find the interval of convergence of

1 P n=0

1 . x2n

3 Find the radius and interval of convergence for each of the following series: 1 1 1 P P P 3n xn (¡1)n x2n¡1 a n5n xn b c d 2 n=0 (n + 1)

n=1

n=1

4 Find the radius and interval of convergence of

(2n ¡ 1)!

1 P 2 £ 4 £ 6 £ :::: £ (2n)xn n=1

1 P

1 £ 3 £ 5 £ :::: £ (2n ¡ 1)

(¡1)n

n=2

(2x + 3)n n ln n

.

5 A function f is defined by f (x) = 1 + 2x + x2 + 2x3 + x4 + ::::, so f is a power series with c2n¡1 = 2 and c2n = 1 for all n 2 Z + . a Find the interval of convergence for the series.

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b Write an explicit formula for f (x).

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88

CALCULUS 1 P

6 Suppose that the radius of convergence of a power series

cn xn

is R.

n=0

1 P

Find the radius of convergence of the power series

cn x2n .

n=0

1 P

7 Suppose the series

1 P

has radius of convergence 2 and the series

cn xn

n=0

dn xn

has radius

n=0

of convergence 3. What can you say about the radius of convergence of the series

1 P

(cn + dn )xn ?

n=0

1 P xn

8 Show that the power series

n=1

1 P nxn¡1

and the series of derivatives

n2 3n

n2 3n

n=1

have the same

radius of convergence, but not the same interval of convergence. µ1 ¶ Z xµ 1 n¶ P xn P t d 9 Find and dt. For what values of x do these series converge? dx

µ1 P

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Z b

dx

x

(¡1)n 2n n=0 x

¡2

¶

2n

¡1:5 µ P 1

¶

Z dx

c 0

1 2

µ

1 P

¶ x2n

dx

n=0

dx defined? Explain your answer.

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1 P xn¡1

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Z

0:1

5

10 Find: Z a

n!

n=1

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CALCULUS

L

89

TAYLOR AND MACLAURIN SERIES 1 P

Let

cn (x ¡ a)n

be a power series with radius of convergence

R > 0,

and interval of

n=0

convergence I. 1 P

For each x with jx ¡ aj < R, the series 1 P

therefore define a function f (x) =

cn (x ¡ a)n

converges to a finite value. The series may

n=0

cn (x ¡ a)n

with domain I.

n=0

Functions defined in this way may look awkward. However, as we have seen, convergent power series can be added, differentiated, and integrated just like ordinary polynomials. They are very useful because we can express many different functions as power series expansions.

TAYLOR SERIES EXPANSIONS Suppose f (x) =

1 P

cn (x ¡ a)n

n=0

= c0 + c1 (x ¡ a) + c2 (x ¡ a)2 + :::: where

jx ¡ aj < R.

Since we can differentiate the power series on an open interval contained in I, f 0 (x) = c1 + 2c2 (x ¡ a) + 3c3 (x ¡ a)2 + :::: f 00 (x) = 2c2 + 6c3 (x ¡ a) + :::: .. . f (n) (x) = n! cn + (n + 1)! cn+1 (x ¡ a) + :::: f (a) = c0

Using the above formulae, we find that:

f 0 (a) = c1 f 00 (a) = 2c2 = 2!c2 .. . f (n) (a) = n!cn Hence cn =

f (n) (a) n!

where 0! = 1 and f (0) (x) = f (x).

We can use this information to reconstruct the function f in terms of the values of its derivatives evaluated at x = a: For such a function f, the Taylor series expansion of f (x) about x = a is 1 P f 0 (a) f 00 (a) f 000 (a) f (k) (a) f (x) = f(a) + (x ¡ a) + (x ¡ a)2 + (x ¡ a)3 :::: = (x ¡ a)k . 1!

2!

3!

k=0

k!

When a = 0, this is also called the Maclaurin series expansion of f (x). This is 1 P x2 x3 f (k) (0)xk f (x) = f(0) + xf 0 (0) + f 00 (0) + f 000 (0) + :::: =

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90

CALCULUS

We have shown for a function f (x) with domain I, that if: 1 f has a convergent power series representation for x 2 I, and 2 the derivatives of f of all orders exist on I, or I with its endpoints removed if I is closed, then f has a Taylor series representation and this is the only power series representation of f. However, we still need to consider the conditions under which the power series representation of f will converge.

TAYLOR POLYNOMIALS If f (k) (x) exists at x = a for k = 0, 1, ...., n, then the nth degree Taylor polynomial approximation to f (x) about x = a is Tn (x) = f (a) + f 0 (a)(x ¡ a) + :::: + =

n P (x ¡ a)k

k!

k=0

f (n) (a) (x ¡ a)n n!

f (k) (a)

When a = 0, this is also called the nth degree Maclaurin polynomial

n P xk k=0

k!

f (k) (0).

Consider the function f(x) = ex . f (n) (x) = ex

exists for all n 2 Z + and x 2 R , and f (n) (0) = e0 = 1 for all n 2 Z + .

The nth degree Taylor approximation to ex about 0 is Tn (x) = 1 +

x x2 xn + + :::: + . 1! 2! n!

Graphs of f(x) = ex , T1 (x) = 1 + x, T2 (x) = 1 + x +

x2 , 2!

T5 (x) = 1 + x +

x2 x3 x4 x5 + + + 2! 3! 4! 5!

y

T2 (x) = 1 +

and

x x2 + 1! 2!

2 1

f(x) = e

x

are shown alongside:

-4

-3

-2

-1

1

2

x

-1

T5 (x) = 1 + x +

x2 x3 x4 x5 + + + 2! 3! 4! 5!

T1 (x) = 1 + x -2

Note that Tn (x) ¼ f (x) when x is close to 0. As n increases, Tn (x) approximates f (x) = ex closely for an increasingly large subset of I = R . If we denote Rn (x : a) to be the error involved in using Tn (x) to approximate f(x) about x = a on I, then f (x) = Tn (x) + Rn (x : a).

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The graphs for the case of f(x) = ex expanded about x = 0 suggest that as n increases, Rn (x : 0) decreases and Tn (x) becomes closer to f(x).

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CALCULUS

91

Consider a function f with domain I containing x = a (not at an endpoint of I). Suppose f (n+1) (x) exists on I, or I with its endpoints removed if I is closed. For x 2 I, Taylor’s formula with remainder is f (x) = Tn (x) + Rn (x : a) n P f (k) (a)(x ¡ a)k where Tn (x) = k!

k=0

and Rn (x : a) =

f (n+1) (c)(x ¡ a)n+1 (n + 1)!

where c is between x and a, so either c 2 ] a, x [ or c 2 ] x, a [ , depending on the values of a and x. This formula for Rn (x : a) is called the Lagrange form of the error term. Proof of Taylor’s Formula: Consider an interval [a, b] µ I. We define a new function f 00 (x) f (n) (x) (b ¡ x)2 ¡ :::: ¡ (b ¡ x)n ¡ K(b ¡ x)n+1 2! n!

h(x) = f (b) ¡ f (x) ¡ f 0 (x)(b ¡ x) ¡

where K is the constant found by solving h(a) = 0. Note that h(b) = 0. Since f and its n + 1 derivatives exist on [a, b], h is continuous on [a, b] and differentiable on ] a, b [ . Since h(a) = h(b) = 0 we can apply Rolle’s theorem to h on [a, b]. Thus there exists c 2 ] a, b [ such that h0 (c) = 0. Using the product rule n times we obtain for x 2 I and x > a: · ¸ £ 0 ¤ f (3) (x)(b ¡ x)2 0 0 00 00 h (x) = ¡f (x) + f (x) ¡ f (x)(b ¡ x) + f (x)(b ¡ x) ¡ 2! · ¸ · ¸ (3) 2 (4) 3 (n) f (x)(b ¡ x) f (x)(b ¡ x) f (x)(b ¡ x)n¡1 f (n+1) (x)(b ¡ x)n + ¡ + :::: + ¡ 2!

3!

(n ¡ 1)!

n!

+ (n + 1)K(b ¡ x)

n

) h0 (x) = ¡

f (n+1) (x)(b ¡ x)n + (n + 1)K(b ¡ x)n n!

Since h0 (c) = 0, (n + 1)K(b ¡ c)n = )

K=

f (n+1) (c) (b ¡ c)n n! f (n+1) (c) (n + 1)!

We can repeat the argument with interval [b, a] µ I, for x 2 I and x < a. It follows that Rn (x : a) =

f (n+1) (c)(x ¡ a)n+1 (n + 1)!

as required.

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Before the invention of sophisticated calculators, calculations involving irrational numbers like e were difficult. Approximations using Taylor polynomials and rational values of x were often used.

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92

CALCULUS

Example 40 a Find the 3rd degree Taylor polynomial of sin x about ¡ ¢ b Hence estimate sin ¼5 .

¼ 4.

c Use the error term to show your approximation is accurate to at least 3 decimal places. f(x) = sin x

)

f 0 (x) = cos x

)

f 00 (x) = ¡ sin x

)

f 000 (x) = ¡ cos x

)

a

f

¡¼¢

= sin

4

¡ ¢ 0 ¼

¡¼¢ 4

¡¼¢

=

p1

2

p1 4 4 = 2 ¡ ¢ ¡ ¢ f 00 ¼4 = ¡ sin ¼4 = ¡ p12 ¡ ¢ ¡ ¢ f 000 ¼4 = ¡ cos ¼4 = ¡ p12

f

= cos

³ ´³ ´2 ³ ´³ ´3 f 00 ¼4 x ¡ ¼4 f 000 ¼4 x ¡ ¼4 ¡ ¢ ¡ ¢ T3 (x) = f 4 + f 0 ¼4 x ¡ ¼4 + + 2! 3! ³ ´2 ³ ´3 ¼ ¼ x¡ 4 x¡ 4 ¡ ¢ = p12 + p12 x ¡ ¼4 ¡ p12 ¡ p12

¡¼¢

)

2!

b T3

¡¼¢ 5

=

p1

2

+

p1

¡¼ 5

2

¡

¢ ¼ 4

³ ¡

p1

2

¼ 0:5878 ¯ ³ ´ ¯ ¯ sin c ¼ ¡ ¼ 4 ¯ ¯ ¯ ¡ ¢¯ ¯ 5 4 ¯ c ¯R3 ¼5 : ¼4 ¯ = ¯¯ ¯ 4! ¯ ¯ ³ ´4 6

)

¼ ¡ 20

¡ ¼4

³ ¡

2!

where

¼ 5

p1

¼ 5

´3 ¡¼ 4

2

3!

0 for all k, then n Q (1 + uk ) = (1 + u1 )(1 + u2 ) :::: (1 + un ) 6 eu1 +u2 +::::+un .

6

k=1

1 P

c If n Q

uk

n Q

converges, deduce the behaviour of

n=1

uk = u1 £ u2 £ :::: £ un

k=1

(1 + uk ) as n ! 1.

k=1

7 In this question, use the following steps for Euler’s proof of

1 P 1 n=1

n2

=

¼2 . 6

x3 x5 x7 You may assume that sin x = x ¡ + ¡ + :::: for all x 2 R . 3! 5! 7! sin x a Find all the zeros of sin x and of for x 2 R . x sin x b Find the power series expansion for and its interval of convergence. x

c Find all the zeros of

³ ´³ ´³ ´³ ´ x x x x 1¡ 1+ 1¡ 1+ :::: ¼

2¼

¼

2¼

d Show that: µ ¶µ ¶µ ¶ ³ ´³ ´³ ´³ ´ x x x x x2 x2 x2 1¡ 1+ 1¡ 1+ :::: = 1 ¡ 2 1¡ 2 1 ¡ 2 :::: ¼

2¼

¼

2¼

4¼

¼

and comment on Euler’s claim that x2 x4 x6 + ¡ + :::: = 1¡ 3! 5! 7!

9¼

µ ¶µ ¶µ ¶ x2 x2 x2 1¡ 2 1¡ 2 1 ¡ 2 :::: 4¼

¼

9¼

e By equating the coefficients of x2 in this last equation, prove that: 1 P 1 1 1 1 1 ¼2 = 2 + 2 + 2 + 2 + :::: = . 2 n=1

1 P 1

f As

n=1

n2

1

n

2

3

4

6

is absolutely convergent, we can write

n=1

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Use this last equation to find the exact values of

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1 P 1

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=

1 P

1 P 1 1 + . 2 (2r) (2r ¡ 1)2 r=1 r=1

| {z } even n 1 P and

|

{z } odd n

1 . (2n ¡ 1)2 n=1


Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\103IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 9:13:17 AM BRIAN

104

CALCULUS

HISTORICAL NOTE Euler was able to derive a method to sum all series of the 1 P 1 form , k 2 Z +. 2k n

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1 P

0

However, the exact value of

5

n=1

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Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_01\104IB_HL_OPT-Calculus_01.cdr Monday, 11 February 2013 9:15:15 AM BRIAN

CALCULUS

M

105

DIFFERENTIAL EQUATIONS

A differential equation is an equation involving the derivative(s) of an unknown function. Suppose y is a function of x, so y = y(x). Examples of differential equations for this function are: dy x2 = dx y

d2 y dy ¡3 + 4y = 0 dx2 dx

dy = ¡0:075y 3 dx

Such equations not only arise in pure mathematics, but are also used to model and solve problems in applied mathematics, physics, engineering, and the other sciences. For example: A parachutist

A falling object

Object on a spring

y

m

v d2 y = 9:8 dx2

m

dv = mg ¡ av2 dt

m

Water from a tank

Current in an RL Circuit

d2 y = ¡ky dt2

Dog pursuing cat y

curve of pursuit

R

(x, y)

E L

L

H

p dH = ¡a H dt

dI + RI = E dt

x

d2 y = dx2

r 1+

x

³

´ dy 2 dx

FIRST ORDER DIFFERENTIAL EQUATIONS In this course we will only deal with differential equations of the form f(x, y)

dy + g(x, y) = 0 dx

where y = y(x).

These are known as first order differential equations since there is only one derivative in the equation, and it is a first derivative.

SOLUTIONS OF DIFFERENTIAL EQUATIONS A function y(x) is said to be a solution of a differential equation if it satisfies the differential equation for all values of x in the domain of y. dy = x. dx

Z

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x dx = 12 x2 + c.

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For example, a very simple differential equation is

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106

CALCULUS

In solving the differential equation we obtain a constant of integration. We say that y = 12 x2 + c is a general solution to the differential equation. It describes a family of curves which all satisfy the differential equation. If we are given initial conditions for the problem, such as a value of y or

dy for a specific value of x, dx

then we can evaluate c. This gives us a particular solution to the problem, which is one particular curve from the family of curves described by the general solution. Example 47 Consider the differential equation

dy ¡ 3y = 3. dx

a Show that y = ce3x ¡ 1 is a solution to the differential equation for any constant c. b Sketch the solution curves for c = §1, §2, §3. c Find the particular solution which passes through (0, 2). d Find the equation of the tangent to the particular solution at (0, 2). a If y = ce3x ¡ 1 then )

dy = 3ce3x . dx

dy ¡ 3y = 3ce3x ¡ 3(ce3x ¡ 1) dx

= 3ce3x ¡ 3ce3x + 3 =3 ) the differential equation is satisfied for all x 2 R .

y

b The solution curves for c = §1, §2, §3 are shown alongside:

y = 2e 3x - 1

2 y = 3e 3x - 1 -1

-2

y = e 3x - 1 1 x

y = -3e

3x

-1 -4

-2 y = -e 3x - 1 y = -2e 3x - 1

c y = ce3x ¡ 1 is a general solution to the differential equation. The particular solution passes through (0, 2), so 2 = ce3£0 ¡ 1 ) c=3 ) the particular solution is y = 3e3x¡1 dy = 3 + 3y dx

d

y

) at the point (0, 2),

dy =3+3£2=9 dx

) the gradient of the tangent to the particular solution y = 3e3x¡1 at (0, 2), is 9. ) the equation of the tangent is

-2

2 y = 3e 3x - 1 -1

x

y¡2 =9 x¡0

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CALCULUS

107

SLOPE FIELDS Given any first order differential equation of the form f (x, y) we can write

dy + g(x, y) = 0 dx

where y = y(x),

dy g(x, y) =¡ = h(x, y). dx f (x, y)

We may therefore deduce the gradient of the solution curves to the differential equation at any point (x, y) in the plane where h(x, y) is defined, and hence the equations of the tangents to the solution curves. The set of tangents at all points (x, y) is called the slope field of the differential equation. For example, the table alongside shows the values of dy = x(y ¡ 1) dx

for the integer grid points x, y 2 [¡2, 2]. y

¡2 ¡1 0 1 2

¡2 6 4 2 0 ¡2

x 0 0 0 0 0 0

¡1 3 2 1 0 ¡1

1 ¡3 ¡2 ¡1 0 1

2 ¡6 ¡4 ¡2 0 2

y

By representing the gradients at many different grid points as line segments, we obtain a slope field of the tangents to the solution curves as shown.

4 2

-4

-2

2

4 x

2

4 x

-2 -4 y

Now the tangent to a curve approximates that curve at and near the points of tangency. Therefore, by following the direction given by the slope field at a given point, we can approximate solution curves of the differential equation. The horizontal line in the figure is the solution curve corresponding to the initial condition that y = 1 when x = 0.

4 2

-4

-2 -2 -4

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Although it is quite straightforward to obtain a few slope field points by hand, a larger or more refined field is best obtained using technology. You can click on the icon on your CD or else download software for your graphics calculator.

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SLOPE FIELDS

108

CALCULUS

An isocline is the set of all points on a slope field at which the tangents to the solution curves have the same gradient. For a first order differential equation written in the form

Isoclines are not solutions to the differential equation. In general, they are not straight lines.

dy = h(x, y), an isocline is a curve given by h(x, y) = k, dx

where k is a particular constant. There may also be an isocline corresponding to points where the gradients of the tangents to the solution curve are undefined.

For example, suppose

dy 1 ¡ x2 ¡ y2 = . dx y¡x+2

dy is discontinuous when dx

²

y 2

y ¡ x + 2 = 0, which is

when y = x ¡ 2. This isocline is the red line on the slope field. dy is zero when dx

²

-2

1 ¡ x2 ¡ y 2 = 0,

2

which is when x2 + y 2 = 1. This isocline is the green circle on the slope field.

x

-2

Example 48 dy = xy dx

Consider the differential equation a Verify that y = ce

x2 2

where x, y > 0.

is a general solution for any constant c 2 R , c > 0.

b Construct the slope field using integer grid points for x, y 2 [0, 4]. c Sketch the particular solution curve through P(2, 1). 2

d Verify algebraically that the curve you have sketched is y = e Draw the isoclines corresponding to k = 1, 3, and 6. a If y = ce

x2 2

1 x2 e . e2

2

then

x dy x = 2 £ £ ce 2 dx 2 x2

= xce 2 µ x2 ¶ = x ce 2 = xy > 0 for all x

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CALCULUS

b, c 0 0 0 0 0 0

0 1 2 3 4

y

x 2 0 2 4 6 8

1 0 1 2 3 4

3 0 3 6 9 12

y 4

4 0 4 8 12 16

2

y=

1 x2 e e2

3 2

k = 6, y =

x2 2

d The general solution is y = ce . 1 The particular solution passes through (2, 1)

k = 3, y =

1 = ce 1 = ce2

)

c=

3 x

k = 1, y =

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6 x

1

2

3

1 x

x

4

1 e2 2

Hence P(2, 1) lies on the particular solution curve y =

1 x2 e . e2

e The isoclines are shown in red.

EULER’S METHOD OF NUMERICAL INTEGRATION Euler’s Method uses the same principle as slope fields to find a numerical approximation to the solution of the differential equation Since the gradient

dy = f (x, y). dx

dy indicates the direction in which the solution curve goes at any point, we reconstruct dx

the graph of the solution as follows: We start at a point (x0 , y0 ) and move a small distance in the direction of the slope field to find a new point (x1 , y1 ). We then move a small distance in the direction of the slope field at this new point, and so on. If we step h units to the right each time, then x1 = x0 + h and y1 = y0 + h f (x0 , y0 ).

gradient¡=¡f(x0, y 0)

(x0, y 0)

More generally, xn+1 = xn + h and yn+1 = yn + h f (xn , yn ).

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Clearly, Euler’s Method only gives an approximate solution to an initial value problem. However, by decreasing the step size h and hence increasing the number of course corrections, we can usually improve the accuracy of the approximation.

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(x1, y 1)

h

Euler’s Method will be less accurate when the gradient f (x, y) is large.



110

CALCULUS

Example 49 dy =x+y dx

Suppose

where y(0) = 1. Use Euler’s Method

with a step size of 0:2 to find an approximate value for y(1). Now xn+1 = xn + h ) given f (x, y) =

If we are given a differential equation and an initial point, we call it an initial value problem or IVP.

and yn+1 = yn + h f(xn , yn ) dy =x+y dx

xn+1 = xn + 0:2

and step size h = 0:2,

and yn+1 = yn + 0:2(xn + yn )

Using the initial conditions, y0 x0 = 0 y1 x1 = 0 + 0:2 = 0:2 y2 x2 = 0:2 + 0:2 = 0:4 y3 x3 = 0:4 + 0:2 = 0:6 y4 x4 = 0:6 + 0:2 = 0:8 y5 x5 = 0:8 + 0:2 = 1

=1 = 1 + 0:2(0 + 1) = 1:2 = 1:2 + 0:2(0:2 + 1:2) = 1:48 = 1:48 + 0:2(0:4 + 1:48) = 1:856 = 1:856 + 0:2(0:6 + 1:856) = 2:3472 = 2:3472 + 0:2(0:8 + 2:3472) = 2:9766

So, y(1) ¼ 2:98 .

EXERCISE M dy = 2x ¡ y. dx

1 Consider the differential equation

a Show that y = 2x ¡ 2 + ce¡x

is a solution to the differential equation for any constant c.

b Sketch the solution curves for c = 0, §1, §2. c Find the particular solution which passes through (0, 1). d Find the equation of the tangent to the particular solution at (0, 1). a Show that y =

2

p x2 + c is a general solution to the differential equation

for any c 2 R . b Hence solve the IVP

dy x = , dx y

dy x = , dx y

y > 0,

y(3) = 4.

3 Slope fields for two differential equations are plotted below for x, y 2 [¡2, 2]. Use the slope fields to graph the solution curves satisfying y(1) = 1. a

b

y

y

2

2

x 2 x

-2

-2

2

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CALCULUS

dy = 10y tan x dx


111

where x is measured in degrees. Draw the

slope field using integer grid points where x, y 2 [¡2, 2]. dy = x2 + y ¡ 1. dx


Sketch isoclines for k = 0, 2, 4 on a grid for x, y 2 [¡5, 5]. Hence construct a slope field, and sketch the solution curve satisfying y(0) = 1. 6 The slope field for the differential equation dy ¡1 + x2 + 4y 2 = dx y ¡ 5x + 10

y

is shown alongside.

2

a Sketch the particular solution passing through the origin.

1 -3

b Sketch the isocline corresponding to: i

dy being undefined dx

1

dy = 0. dx

ii

-1 3

x

-1 -2

7 Use Euler’s Method with step size 0:2 to estimate y(1) for the initial value problem dy = 1 + 2x ¡ 3y, dx

y(0) = 1.

8 Use Euler’s Method with step size 0:1 to estimate y(0:5) for the initial value problem

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y(0) = 0:5. Assume x and y are in radians.

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dy = sin(x + y), dx

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112

CALCULUS

N

SEPARABLE DIFFERENTIAL EQUATIONS dy f (x) = , dx g(y)

Differential equations which can be written in the form

where y = y(x), are known as

separable differential equations. dy f (x) = dx g(y)

Notice that if

then g(y)

dy = f(x). dx

If we integrate both sides of this equation with respect to x we obtain Z Z dy g(y) dx = f (x) dx dx Z Z ) by the Chain Rule, g(y) dy = f (x) dx. The problem of solving the differential equation is hence reduced to the problem of finding two separate integrals. Example 50 Solve the initial value problem 2x

dy ¡ 1 = y2 , dx

dy ¡ 1 = y2 dx dy ) 2x = y2 + 1 dx 1 1 dy = y 2 + 1 dx 2x

y(1) = 1.

2x

)

Z

Integrating both sides with respect to x gives

1 dy dx = y 2 + 1 dx

Z

)

1 dy = y2 + 1

)

arctan y = )

¡1

¢ But y(1) = 1, so 1 = tan 2 ln 1 + c ) 1 = tan c ) c = ¼4

Z Z

1 dx 2x 1 dx 2x

1 2

ln jxj + c ¡ ¢ y = tan 12 ln jxj + c

) the particular solution of the differential equation is y = tan

¡1 2

ln jxj +

¼ 4

¢ .

Example 51 a Show that

2 1 1 = ¡ . x2 ¡ 1 x¡1 x+1

b Find the general solution of the differential equation

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1 1 x + 1 ¡ (x ¡ 1) 2 ¡ = = 2 as required. x¡1 x+1 (x ¡ 1)(x + 1) x ¡1

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dy x2 y + y = 2 . dx x ¡1

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CALCULUS

dy x2 y + y = 2 dx x ¡1

b

1 dy x2 + 1 = 2 y dx x ¡1 x2 ¡ 1 + 2 = x2 ¡ 1 2 =1+ 2 x ¡1 1 1 =1+ ¡ x¡1 x+1

)

y(x2 + 1) x2 ¡ 1

=

113

fusing ag

Integrating both sides with respect to x gives Z Z ³ ´ 1 dy 1 1 dx = 1+ ¡ dx y dx x¡1 x+1 Z 1 ) dy = x + ln jx ¡ 1j ¡ ln jx + 1j + c y ¯´ ³ ¯ ¯x ¡ 1¯ where ln A = c ln jyj = x + ln A ¯ ¯ x+1 ³ ´ x¡1 is the general solution of the differential equation. ) y = Aex x+1

Example 52 When an object travels through a resistive medium, the rate at which it loses speed at any given instant is given by kv m s¡2 , where v is the speed of the body at that instant and k is a positive constant. Suppose the initial speed is u m s¡1 . By formulating and solving an appropriate differential equation, show that the time taken for the body to decrease its speed to The rate of change of speed is given by

m s¡1 is

1 ln 2 seconds. k

dv . dt

Our differential equation must reflect that the body loses speed, so 1 dv = ¡k. v dt

Separating the variables, we find

1 2u

Z

Integrating both sides with respect to t gives )

1 dv = ¡k v

dv = ¡kv. dt

Z dt

ln jvj = ¡kt + c v = Ae¡kt

)

where A = ec

The initial speed (at t = 0) is u, so u = Ae¡k£0 = A. ) v = ue¡kt is the particular solution of the differential equation. When v = 12 u we have )

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114

CALCULUS

Example 53 Consider a curve in the first quadrant. The tangent at any point P cuts the x-axis at Q. Given that OP = PQ, where O is the origin, and that the point (1, 4) lies on the curve, find the equation of the curve. Since OP = OQ, triangle OPQ is isosceles. Hence [PA] is the perpendicular bisector of [OQ].

y P(x, y)

The coordinates of A are (x, 0), so the coordinates of Q are (2x, 0). Since (PQ) is the tangent to the curve at P, the gradient of the curve at P is the same as the gradient of [PQ]. Hence )

dy y =¡ dx x 1 dy 1 =¡ y dx x

Z

Integrating both sides with respect to x gives ) )

O

A x

1 dy = ¡ y

Z

x

Q x

1 dx x

ln jyj = ¡ ln jxj + c

ln jxj + ln jyj = c )

ln jxyj = c xy = ec = k

)

where k is a constant.

Since the curve passes through (1, 4), 1 £ 4 = k ) the equation of the curve is xy = 4 or y =

4 , where x > 0. x

HOMOGENEOUS DIFFERENTIAL EQUATIONS ³ ´

Differential equations of the form

dy y = f , dx x

where y = y(x), are known as homogeneous

differential equations. They can be solved using the substitution y = vx where v is a function of x. The substitution will always reduce the differential equation to a separable differentiable equation as follows. If y = vx where v is a function of x, then

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dy dv = x + v fproduct ruleg dx dx ³ ´ dv vx x+v =f = f(v) ) dx x dv f (v) ¡ v = ) dx x 1 1 dv ) = which is a separable differential equation. f (v) ¡ v dx x

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CALCULUS

115

Example 54 dy x + 2y = . dx x

a Use the substitution y = vx, where v is a function of x, to solve b Find the particular solution if y =

3 2

when x = 3.

dy dv =v+x . dx dx dv x + 2vx = Comparing with the differential equation, we find v+x dx x dv ) v+x = 1 + 2v dx dv ) x =1+v dx 1 dv 1 ) = 1 + v dx x

a If y = vx, then using the product rule we obtain

Z

Integrating with respect to x gives ) ) But v =

y , so x

1 dv = v+1

Z

1 dx x

ln jv + 1j = ln jxj + c ln jv + 1j = ln jAxj where ln jAj = c ) v + 1 = Ax

y + 1 = Ax x

y = Ax2 ¡ x, where A is a constant.

) b Substituting y =

3 2

3 2

and x = 3 into the general solution, we find )

9A =

) ) the particular solution is y =

1 2 2x

= A £ 32 ¡ 3

A=

¡ x.

9 2 1 2

EXERCISE N 1 Solve the following initial value problems: a (2 ¡ x)

dy = 1, dx

y(4) = 3

c ey (2x2 + 4x + 1) d x 2

dy = cos2 y, dx

b

dy = (x + 1)(ey + 3), dx

y(e) =

y(1) = 0

y(0) = 2

¼ 4

3¡x 1 2 = ¡ . x2 ¡ 1 x¡1 x+1

a Show that

dy ¡ 3x sec y = 0, dx

b Solve

dy 3y ¡ xy = 2 , dx x ¡1

y(0) = 1.

3 According to Newton’s law of cooling, the rate at which a body loses temperature at time t is proportional to the amount by which the temperature T (t) of the body at that instant exceeds the temperature R of its surroundings.

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a Express this information as a differential equation in terms of t, T , and R. b A container of hot liquid is placed in a room with constant temperature 18± C. The liquid cools from 82± C to 50± C in 6 minutes. Show that it takes 12 minutes for the liquid to cool from 26± C to 20± C.

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CALCULUS

4 The tangent at any point P on a curve cuts the x-axis at the point Q. PQ = 90± , where O is the origin, and that the point (1, 2) lies on the curve, find Given that Ob the equation of the curve. 5 The tangent at any point P on a curve cuts the x-axis at A and the y-axis at B. Given that AP : PB = 2 : 1 and that the curve passes through (1, 1), find the equation of the curve. 6 A radioactive substance decays at a rate proportional to the mass m(t) remaining at the time. Suppose the initial mass is m0 . a Construct and solve the appropriate initial value problem and hence obtain a formula for m(t). b If the mass is reduced to 45 of its original value in 30 days, calculate the time required for the mass to decay to half its original value. 7 Solve the following homogeneous differential equations using the substitution y = vx, where v is a function of x. a

dy x¡y = dx x

b

dy x+y = dx x¡y

dx

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y dy x = y + ex dx

b Solve

0

dy y 2 ¡ x2 = dx 2xy

a Show that the substitution y = vx (where v is a function of x) will reduce all inhomogeneous ³ ´ dy y y = +f g(x) to separable form. differential equations of the form

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CALCULUS

O

THE INTEGRATING FACTOR METHOD

Suppose a first order linear differential equation is of the form

dy +P (x)y = Q(x), dx

where y = y(x).

Generally, but not always, this type of equation is not separable. However, suppose there is a function I(x), called an integrating factor, such that d dy (I(x)y) = I(x) + I(x) P (x)y dx dx

.... (¤)

= I(x) Q(x) Then integrating both sides with respect to x would give Z I(x)y = I(x) Q(x) dx Z 1 I(x) Q(x) dx and we could hence find a solution for y. ) y= I(x)

Now if such an integrating factor exists, then from (¤), I(x)

dy dy + I 0 (x)y = I(x) + I(x) P (x)y dx dx

)

I 0 (x) = I(x) P (x)

)

I 0 (x) = P (x) I(x)

Integrating both sides with respect to x, Z 0 Z I (x) dx = P (x) dx I(x) Z ln jIj + c = P (x) dx R ) I(x) = Ae P (x) dx where A = e¡c and is conventionally set as 1. R I(x) = e

Thus the integrating factor is

P (x) dx

.

THE INTEGRATION FACTOR METHOD dy + P (x) y = Q(x) dx

Suppose

where y = y(x).

R 1 Calculate the integrating factor I(x) = e integration.

P (x) dx

. You do not need a constant of

2 Multiply the differential equation through by I(x). Z 3 Simplify the LHS and hence obtain I(x) y = I(x) Q(x) dx + c, where c is a constant.

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4 Integrate to obtain the general solution.

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118

CALCULUS

Example 55 Solve the differential equation

dy + 3x2 y = 6x2 . dx

R The integrating factor is I(x) = e

3x2 dx

= ex

3 3

Multiplying the differential equation through by ex gives 3

3 3 dy + 3x2 ex y = 6x2 ex dx ³ 3´ 3 d ) yex = 6x2 ex dx Z

ex

3

)

3

)

3

6x2 ex dx

yex =

3

yex = 2ex + c 3 ) y = 2 + ce¡x

Example 56 Solve the initial value problem cos x

dy = y sin x + sin(2x), dx

y(0) = 1.

dy sin x sin(2x) ¡ y= dx cos x cos x

We can rewrite the differential equation as

dy + (¡ tan x)y = 2 sin x dx

)

The differential equation is not separable, but is of a form such that we can use an integrating factor. R The integrating factor is I(x) = e ¡ tan x dx = eln(cos x) = cos x. Multiplying the equation through by the integrating factor gives cos x

dy + (¡ cos x tan x)y = sin(2x) dx d (y cos x) = sin(2x) ) dx Z

)

y cos x =

sin(2x) dx

= ¡ 12 cos(2x) + c But when x = 0, y = 1 ) 1 = ¡ 12 cos(0) + c ) c = 32

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CALCULUS

119

EXERCISE O 1 Solve the following using the integrating factor method. a

dy + 4y = 12 dx

c

dy + y = x + ex , dx

b y(1) = 1

dy ¡ 3y = ex , dx

d x

2 Solve the differential equation (x + 1)y + x

y(0) = 2

dy + y = x cos x dx

dy = x ¡ x2 . dx

ACTIVITY

LAPLACE TRANSFORMS

Laplace transforms provide a useful link between improper integrals and differential equations. Z 1 e¡sx f (x) dx. The Laplace transform of a function f (x) is defined as F (s) = Lff(x)g = 0

What to do: 1 Show that: a Lfeax g =

1 , s¡a

c Lfsin(ax)g =

b Lfxg =

s>a

a , s2 + a2

1 , s2

s>0

s>0

2 Show that: a Lff 0 (x)g = sLff(x)g ¡ f (0)

b Lff 00 (x)g = s2 Lff(x)g ¡ sf(0) ¡ f 0 (0)

3 Consider the differential equation f 00 (x) + f (x) = x, f (0) = 0, f 0 (0) = 2. Assuming that Lfg(x)+h(x)g = Lfg(x)g + Lfh(x)g, show that Lff(x)g =

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120

CALCULUS

P

TAYLOR OR MACLAURIN SERIES DEVELOPED FROM A DIFFERENTIAL EQUATION

Consider a first order differential equation with initial condition: dy = h(x, y) dx

where y(x0 ) = y0 for some constants x0 , y0 2 R .

Depending on the form of h(x, y), it may or may not be possible to obtain an explicit particular solution y(x). However, provided the necessary derivatives exist, we can determine the nth degree Taylor polynomial approximation to the particular solution y(x) about x0 as follows: y(x0 ) = y0 y0 (x) = h(x, y) d (h(x, y)) dx ¢ d ¡ 00 y (x) :::: y000 (x) = dx

y00 (x) =

)

y 0 (x0 ) = h(x0 , y0 )

)

we can obtain y 00 (x0 ) using x0 , y0 , and y 0 (x0 ) found above.

Once y (x0 ), y 0 (x0 ), y00 (x0 ), ::::, y(n) (x0 ) have been determined, the nth degree Taylor polynomial approximation to y(x) about x0 can then be written as: Tn (x) = y (x0 ) + y0 (x0 ) (x ¡ x0 ) +

y00 (x0 ) (x ¡ x0 )2 y(n) (x0 ) (x ¡ x0 )n + :::: + . 2! n!

Example 57 Consider the initial value problem

dy = x ¡ 4xy, dx

y(0) = 1.

a Without solving the system, find the first three non-zero terms of the Maclaurin polynomial expansion for y. b Solve the system exactly to find the particular solution.

)

dy = x ¡ 4xy = x (1 ¡ 4y) dx d2 y d = (x (1 ¡ 4y)) dx2 dx ³ ´ dy = 1 ¡ 4y + x ¡4 dx

)

d3 y dy dy = ¡4 + 1 £ ¡4 dx3 dx dx

³

d4 y d2 y d2 y = ¡8 2 ¡ 4 2 + x 4 dx dx dx

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µ ¶ d2 y + x ¡4 2 dx

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dy ¡ 4x 2 dx dx

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´

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CALCULUS

121

dy =0 dx

Now y(0) = 1, so at (0, 1):

d2 y = 1 ¡ 4 + 0 = ¡3 dx2 d3 y =0 dx3 d4 y = 36 dx4

) y has Maclaurin polynomial 1 ¡

3x2 36x4 + :::: which is 1 ¡ 32 x2 + 32 x4 :::: 2! 4!

dy = x (1 ¡ 4y) dx 1 dy =x 1 ¡ 4y dx

b

Z

1 dy = 1 ¡ 4y

)

¡ 14 ln j1 ¡ 4yj =

Z

x dx But y(0) = 1

x2 +c 2

1 4

)

2

ln j1 ¡ 4yj = ¡2x + c

B = ¡ 34 µ ¶ 3 ) y = 14 1 + 2x2

)

¡2x2

1 ¡ 4y = Ae y=

1 4

¡B = 1

¡ Be¡2x

2

e

EXERCISE P dy e¡y = ¡ 1, dx 3


where y(0) = 0.

a Find the first three non-zero terms in the Maclaurin polynomial expansion of y. µ ¶ 1 + 2e¡x b Verify that y = ln is a particular solution. 3

dy y = 2x + , dx x


where

dy = 1 when x = 1. dx

a Find the first three non-zero terms in the Maclaurin polynomial expansion of y about x = 1. b Solve the system exactly to find the particular solution. dy 3x ¡ 2y = , dx x

3 Consider the initial value problem

y(1) = 0.

a Without solving the system, find the first three non-zero terms of the Taylor polynomial expansion for y about x = 1. b Solve the system exactly to find the particular solution. 4 Consider the initial value problem cos x

dy + y sin x = 1, dx

y(0) = 2.

a Find the first three non-zero terms in the Maclaurin polynomial expansion of y.

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b Solve the system exactly to find the particular solution.

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122

CALCULUS

Example 58 1 P xk

Given that

k!

k=0

steps.

is convergent for all x 2 R , show that ex =

1 k P x k=0

a Show that if y(x) =

1 P xk k=0 k!

c Hence prove ex =

1 P xk

k!

k=0

using the following

dy = y. dx

, then

b Solve the differential equation

k!

dy = y to obtain the general solution. dx

.

1 P xk

x2 x3 =1+x+ + + :::: k! 2! 3! k=0 dy 2x 3x2 4x3 =1+ + + + :::: dx 2! 3! 4!

a Let y(x) = )

=1+x+ =

1 P xk k=0

x2 x3 + + :::: 2! 3!

k!

=y b Consider ) Z )

dy =y dx 1 dy =1 y dx

Z

1 dy = y

1 dx

) ln jyj = x + c, where c is a constant ) y = Aex , where A = §ec is a constant. 1 P xk c Since y(x) = is a solution to the differential equation for all x 2 R , 1 P xk

y(x) =

k!

k=0

k=0

k!

= Aex for some constant A.

Now y(0) = 1, so A = 1 1 P xk for all x 2 R . Hence ex = k=0

5 Let y(x) =

k!

1 P p(p ¡ 1) :::: (p ¡ n + 1)xn

n!

n=0

for jxj < 1. From Exercise L.2 question 8, y is a well

defined function on jxj < 1 since the binomial series was shown to be convergent on the interval ¡1 < x < 1. a Show that for ¡1 < x < 1, y satisfies the differential equation (1 + x)

dy = py. dx

b Find the general solution to the differential equation in a.

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c Hence show that y(x) = (1 + x)p for all x 2 ] ¡1, 1 [ .

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CALCULUS

123

REVIEW SET A lim

1 Prove that 2 Find

x!1

ln x = 0. x

ex sin x . x!0 x

lim

3 Find the limit, if it exists, of the sequence fun g as n tends to infinity, if un equals: 8 ¡ 2n ¡ 2n2 4 + 6n + 7n2

a

c p

b 3+

2n + 13

d arctan n

6n2 + 5n ¡ 7

4 Prove that the series

5

13

1 2 3 4 5 + 3 + 3 + 3 + 3 + :::: +1 2 +1 3 +1 4 +1 5 +1

ń 1 ³ P 3x

a For what values of x does b Find

in terms of x.

x¡2

1 P n(3x)n¡1

c Show that

n=0

converges.

converge?

x¡2

n=1

ń 1 ³ P 3x n=0

1 + n [1 + (¡1)n ] n

(x ¡ 2)n+1

=

1 4(x + 1)2

for the x values found in a.

6 Find the set of real numbers for which the series x +

Z

1

7 Using an appropriate Maclaurin series, evaluate

x2 x3 + + :::: 1¡x (1 ¡ x)2

converges.

sin(x2 ) dx to three decimal places.

0

8 Let X be a random variable such that P(X = x) = 1 P

Prove that

e¡¸ ¸x x!

for x = 0, 1, 2, ....

P(X = x) = 1.

x=0

a Draw the slope field using integer grid points for x, y 2 [¡3, 3] for the differential dy x = . dx y

b Draw on your slope field the isoclines dy xy = dx x¡1

10 Solve the differential equation

dy =k dx

p dy y ¡ = x, dx x

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11 By finding a suitable integrating factor, solve

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given that y = 2 when x = 2.

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y(4) = 0.



124

CALCULUS

12 On the slope field for

dy = 2x ¡ y 2 dx

y

shown,

sketch the solution curves through a (0, 0)

b (2, 3).

4

2

-2

2

4

x

-2

REVIEW SET B 1 Find the limit, if it exists, of the sequence fun g as n tends to infinity, if un equals: a c

(¡1)n (2n ¡ 1) n

p p n+5¡ n¡1

2 Prove that the series x +

x2 x3 x4 + + + :::: 2 3 4

b

0:9n 1 + 0:1n

d

n2 2n3 ¡ 2 3n + 1 6n + 1

is convergent for ¡1 < x < 1 and divergent

for jxj > 1. Determine the convergence or divergence of the series for x = §1. ³ ´ 1 P (k ¡ 1)¼ sin is convergent. 3 Determine whether or not the series 2k

k=1

1 P 1+r

4 Use the Comparison Test to prove that the series

1 + r2

r=1

a Show that the series Sn =

5

n P (¡1)k+1 k=3

ln(k ¡ 1)

diverges.

converges as n ! 1.

b Find the maximum possible error in using S10 to estimate

1 P (¡1)k+1 k=3

ln(k ¡ 1)

.

6 Find the Taylor series expansion of (x ¡ 1)ex¡1 about x = 1 up to the term (x ¡ 1)3 . 7 Find constants a and b such that y = ax + b is a solution of the differential equation dy = 4x ¡ 2y. dx

8 Find the general solution of the differential equation

dy = 2xy2 ¡ y 2 . dx

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9 Find the equation of the curve through (2, 1) given that for any point (x, y) on the curve, the y-intercept of the tangent to the curve is 3x2 y3 .

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CALCULUS

125

10 Match the slope fields A, B, and C to the differential equations: dy =y+1 dx

a A

dy =x¡y dx

b B

y

c C

y

2

dy = x ¡ y2 dx

y 2

2

2 x

-2

2 x

-2 -2

-2

2 x

-2 -2

11 The population P of an island is currently 154. The population growth in the foreseeable future ³ ´ dP P = 0:2P 1 ¡ , t > 0. is given by 400

dt

1 1 + P 400 ¡ P

a Write

as a single fraction.

b Find P as a function of time t years. c Estimate the population in 20 years’ time. d Is there a limiting population size? If so, what is it? 12 Let f(x) =

1 P

x2 . (1 + x2 )i i=1

a Find f (0). b For x 6= 0, show that

1 P (1 + x2 ) 1 f (x) = ri where r = . 2 x 1 + x2 i=0

c For which values of x 2 R is f(x) convergent? Hence state the domain of f . d For each x for which f(x) is convergent, find the value of f (x). e Sketch y = f (x).

REVIEW SET C 1 Find the limit, if it exists, of the sequence fun g as n tends to infinity, if un equals: 1 p en b (3n + 2n ) n c d (¡1)n ne¡n a n ¡ n2 + n n!

1 P

2 Explain why the series

1

3r

is not convergent.

r=1

1 P

3 Determine whether

n=2

4 Suppose

1 P

a Prove that

is convergent or divergent.

is convergent, where an > 0 for all n 2 Z + .

an

n=1

1 ln(n2 )

1 P

an2

´ 1 ³ P 1 2 an ¡ are also convergent.

and

n=1

n

n=1

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b Would these results follow if an 2 R ?

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126

CALCULUS

5

a Show that

1 1 1 = ¡ . x(x + 1) x x+1

1 P

b Use a and the Integral Test to prove that the series

n=1

1 n(n + 1)

converges.

n P

1 be the nth partial sum for the series in b. i(i + 1) i=1 1 i Use a to show that Sn = 1 ¡ . n+1

c Let Sn =

ii Hence find the value of the sum of the convergent series

1 P

1 . n(n + 1) n=1

6 Let Rn be the error term in approximating f(x) = ln(1 + x) for 0 6 x < 1 using the first n + 1 terms of its Maclaurin series. Prove that jRn j 6

1 n+1

for 0 6 x < 1.

7 Find a simplified expression for 1 ¡ x + x2 ¡ x3 + :::: where ¡1 < x < 1. Hence find a power series expansion for f (x) = ln(1 + x) for ¡1 < x < 1. 8 A curve passes through the point (1, 2) and satisfies the differential equation dy = x ¡ 2y. dx

Use Euler’s Method with step size 0:1 to estimate the value of y when x = 1:6 . 9 A water tank of height 1 m has a square base 2 m £ 2 m. When a tap at its base is opened, the water flows out at a rate proportional to the square root of the depth of the water at any given time. Suppose the depth of the water is h m, and V is the volume of water remaining in the tank after t minutes. a Write down a differential equation involving

dV and h. dt

b Explain why V = 4h m3 at time t. Hence write down a differential equation involving dh and h. dt

c Initially the tank is full. When the tap is opened, the water level drops by 19 cm in 2 minutes. Find the time it takes for the tank to empty. 10 Use the substitution y = vx where v is a function of x, to solve the differential equation dy x y = + . dx y x

11 Use an integrating factor to solve

dy = cos x ¡ y cot x, dx

y

¡¼¢ 2

= 0.

REVIEW SET D 1 Find the limit, if it exists, of the sequence fun g as n tends to infinity, if un equals: ³ ´ ³ ´ 3 £ 5 £ 7 £ :::: £ (2n + 1) 1 1 b n(2 cos ¡ sin ¡ 2) a 2 £ 5 £ 8 £ :::: £ (3n ¡ 1)

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n

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1 n(ln n)2 n=2

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2 Prove that the series

n

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CALCULUS 1 P (x ¡ 3)n

3 Determine the interval and radius of convergence of the series

converges or diverges.

n+5

n=0

.

n2

n=1

ń 1 ³ P n

4 Determine whether the series

3

127

5 Estimate e0:3 correct to three decimal places using the Maclaurin approximation: f 00 (0) x2 f (n) (0) xn f (n+1) (c) xn+1 + :::: + + 2! n! (n + 1)!

f(x) = f (0) + f 0 (0) x +

6 Use the substitution y = vx where v is a function of x, to solve the initial value problem xy 7

dy = 1 + x + y2 , dx

a Prove that e ¡

y(1) = 0. n P 1 k=0

k!

=

ec (n + 1)!

where 0 < c < 1.

b Using the fact that e < 3, show that for n > 3: n P 1 1 3 <e¡ < and hence i (n + 1)!

k=0 n P

(n + 1)!

k!

1 n! 3 < n!e ¡ < 6 34 n+1 k! n + 1 k=0

ii

c Hence, prove by contradiction that e is an irrational number. dy 3y + = 8x4 , dx x

8 By finding a suitable integrating factor, solve y

9

The tangent to a curve at the general point P(x, y)

y = f(x)

3y 2

y(1) = 0.

3y . 2

has x-intercept 3x and y-intercept

P(x, y)

Given that x > 0, find the equation of the curve which passes through the point (1, 5). 3x x

10 The inside surface of y = f (x) is a mirror. Light is emitted from O(0, 0). All rays that strike the surface of the mirror are reflected so that they emerge parallel to the axis of symmetry (the x-axis).

y y = f(x)

P(x, y) ® q

a Explain why µ = 2®. b Explain why the gradient of the tangent at a general point P(x, y) on the mirror is given

x tangent at P

dy = tan ®. dx

by

c Use the identity

y = -f(x)

p

2 tan ® tan(2®) = 1 ¡ tan2 ®

x2 + y 2 ¡ x . y

to deduce that tan ® = p

d Find the general solution to the differential equation the substitution r2 = x2 + y2 .

dy = dx

x2 + y2 ¡ x y

by making

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e What is the nature of y = f (x)?

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CALCULUS


dy = y ln x, x > 0, where y(1) = 1. dx

a Find the first three non-zero terms in the Taylor polynomial expansion of y about x = 1. b Hence estimate the solution to the initial value problem

dy = y ln x, dx

y(1) = 1.

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c Solve the initial value problem in b exactly.

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CALCULUS

THEORY OF KNOWLEDGE Consider f (x) =

129

TORRICELLI’S TRUMPET

1 . An infinite area sweeps out a finite volume. Can this be reconciled with our x

intuition? What does this tell us about mathematical knowledge? Calculus provides us with a set of procedures for manipulating symbols, such as differentiation and integration. Justifying how, and importantly when, such manipulations lead to correct answers is the subject of analysis which underpins calculus. These justifications need to make use of arguments involving infinite processes such as infinitely small quantities and limits over an infinitely large range. These are very subtle concepts which mathematicians and philosophers have discussed and debated for many thousands of years, such as in Archimedes’ work in finding areas and volumes. Much of the debate has focused on paradoxes. A paradox is an argument that appears to produce an inconsistency, such as resulting in 1 = 0. Often the paradox enables a definition, technique, or logical argument to be refined. However there are still some things which appear to be paradoxical and Torricelli’s trumpet, also sometimes called Gabriel’s Horn, is one of these.

To create this shape we take the graph of y =

1 x

for x > 1 and rotate it around the x-axis to

create a solid of revolution. The two questions we wish to ask are: ² What is the volume of this shape? ² What is the surface area of this shape? To answer these questions we shall use the modern techniques of calculus. Torricelli’s original arguments, made in the early 1600s, predate the systematic development of calculus by Newton and Leibnitz later that century. His argument relied on Cavalieri’s Principle: Take two solids included between parallel planes, and cut the solids with another parallel plane. If the areas are always in the same ratio, then the volumes are also in the same ratio.

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Think of a solid, such as a cone, made up of a stack of cards. You can displace the cards without altering the volume. If you can change the shape of the cards without altering the areas then you also don’t change the volume. Cavalieri’s Principle is another extension to the case when the respective areas are in constant ratio.

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CALCULUS

1 What place does an “intuitive” rule such as Cavalieri’s Principle have in mathematics? 2 How can Cavalieri’s Principle be used to find the volume of a sphere from the known volume of a circular cone? Can this be regarded as a valid proof of the formula? In order to consider the volume and surface area we will look at a finite part of the trumpet from x = 1 to x = N and then let N ! 1. VOLUME y

We imagine a plane perpendicular to the x-axis, cutting the axis at x. 1 . x

The trumpet and the plane intersect at a circle of radius The area of the circle is simply ¼

y = Qv

1 . x2

x

To calculate the volume we sum the infinitesimal areas of these Z N h iN ³ ´ 1 1 1 discs as ¼ dx = ¼ ¡ = ¼ 1 ¡ . 2 1

x

x 1

N

As N ! 1 we see that the volume converges to ¼. It is somewhat surprising that an infinite shape can have a finite volume. However, this is similar to the idea that an infinite series of positive terms can sum to a finite length. SURFACE AREA

Z

r

N

To calculate the surface area we use the formula S = 2¼

y

1+

³

1

1 dy 1 we have = ¡ 2 and so the integral becomes Since y = x dx x

Noting that

1 x

Z S = 2¼

r 1 1 1 + 4 > , we find that x

´ dy 2 dx. dx

1

N

1 x

r 1 1 + 4 dx. x

x

Z

N

S = 2¼ 1

1 x

r Z 1 1 + 4 dx > 2¼ x

1

N

1 dx = 2¼ ln(N). x

As N ! 1, ln(N) diverges, and so the surface area S also diverges. Hence we have a finite volume with an infinite surface area. This is a strange and controversial result. Thomas Hobbes, the English Philosopher, said “to understand this for sense, it is not required that a man should be a geometrician or logician, but that he should be mad ”. 3 Imagine filling the trumpet with a liquid. While the trumpet holds a finite volume, what happens when a thin layer coats the surface?

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4 What do we do when a physical problem yields a mathematical result with no physical meaning?

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APPENDICES

APPENDIX A:

131

METHODS OF PROOF

Greek mathematicians more than 2000 years ago realised that progress in mathematical thinking could be brought about by conscious formulation of the methods of abstraction and proof. By considering a few examples, one might notice a certain common quality or pattern from which one could predict a rule or formula for the general case. In mathematics this prediction is known as a conjecture. Mathematicians love to find patterns, and try to understand why they occur. Experiments and further examples might help to convince you that the conjecture is true. However, problems will often contain extra information which can sometimes obscure the essential detail, particularly in applied mathematics. Stripping this away is the process of abstraction. For example, by considering the given table of values one may conjecture: “If a and b are real numbers then a < b implies that a2 < b2 .” However, on observing that ¡2 < 1 but (¡2)2 6< 12 we have a counter-example. In the light of this we reformulate and refine our conjecture:

a 1 3 4 5 6

b 2 5 5 7 9

a2 1 9 16 25 36

b2 4 25 25 49 81

“If a and b are positive real numbers then a < b implies a2 < b2 .” The difficulty is that this process might continue with reformulations, counter-examples, and revised conjectures indefinitely. At what point are we certain that the conjecture is true? A proof is a flawless logical argument which leaves no doubt that the conjecture is indeed a truth. If we have a proof then the conjecture can be called a theorem. Mathematics has evolved to accept certain types of arguments as valid proofs. They include a mixture of both logic and calculation. Generally mathematicians like elegant, efficient proofs. It is common not to write every minute detail. However, when you write a proof you should be prepared to expand and justify every step if asked to do so. We have already examined in the HL Core text, proof by the principle of mathematical induction. Now we consider other methods.

DIRECT PROOF In a direct proof we start with a known truth and by a succession of correct deductions finish with the required result. Example 1: Prove that if a, b 2 R then a 0g fadding

a to both sidesg 2

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Sometimes it is not possible to give a direct proof of the full result and so the different possible cases (called exhaustive cases) need to be considered and proved separately.

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132

APPENDICES

Example 2: Prove the geometric progression: For n 2 Z , n > 0, 8 < rn+1 ¡ 1 , r 6= 1 1 2 n 1 + r + r + :::: + r = r¡1 : n + 1, r=1 1 + r1 + r2 + :::: + rn

Proof: Case r = 1:

= 1 + 1 + 1 + :::: + 1

fn + 1 timesg

=n+1 Let Sn = 1 + r1 + r2 + :::: + rn .

Case r 6= 1:

Then rSn = r1 + r2 + r3 + :::: + rn+1 ) rSn ¡ Sn = rn+1 ¡ 1 ) (r ¡ 1)Sn = r ) Sn =

n+1

fafter cancellation of termsg

¡1

¡1 r¡1

rn+1

fdividing by r ¡ 1 since r 6= 1g

Example 3: Alice looks at Bob and Bob looks at Clare. Alice is married, but Clare is not. Prove that a married person looks at an unmarried person. Proof: We do not know whether Bob is married or not, so we consider the different (exhaustive) cases: If Bob is married, then a married person (Bob) looks at an unmarried person (Clare).

Case: Bob is married.

Case: Bob is unmarried. If Bob is unmarried, then a married person (Alice) looks at an unmarried person (Bob). Since we have considered all possible cases, the full result is proved.

EXERCISE p I 1 Let I = 2, which is irrational. Consider I I and I I , and hence prove that an irrational number to the power of an irrational number can be rational.

PROOF BY CONTRADICTION (AN INDIRECT PROOF)

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In proof by contradiction we deliberately assume the opposite to what we are trying to prove. By a series of correct steps we show that this is impossible, our assumption is false, and hence its opposite is true.

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APPENDICES

133

Example 4: Consider Example 1 again but this time use proof by contradiction: Prove that if a, b 2 R then a 

fmultiplying both sides by 2g

) 2a > a + b ) a>b fsubtracting a from both sidesg which is false. Since the steps of the argument are correct, the supposition must be false and the alternative, a
0. )

x=

p q

where p, q 2 Z , q 6= 0 fand since x > 0, integers p, q > 0g

p q

3 =8 µ p ¶q ) 3q = 8q

)

) 3p = 8q which is impossible since for the given possible values of p and q, 3p is always odd and 8q is always even. Thus, the assumption is false and its opposite must be true. Hence x is irrational. Example 6: Prove that no positive integers x and y exist such that x2 ¡ y2 = 1. Proof (by contradiction): Suppose x, y 2 Z + exist such that x2 ¡ y 2 = 1. ) (x + y)(x ¡ y) = 1 ) x + y = 1 and x ¡ y = 1 or x + y = ¡1 and x ¡ y = ¡1 | {z } | {z } case 1 case 2 ) x = 1, y = 0 (from case 1) or x = ¡1, y = 0 (from case 2) Both cases provide a contradiction to x, y > 0. Thus, the supposition is false and its opposite is true. There do not exist positive integers x and y such that x2 ¡ y 2 = 1.

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Indirect proof often seems cleverly contrived, especially if no direct proof is forthcoming. It is perhaps more natural to seek a direct proof for the first attempt to prove a conjecture.

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134

APPENDICES

ERRORS IN PROOF One must be careful not to make errors in algebra or reasoning. Examine carefully the following examples. Example 7: Consider Example 5 again: Prove that the solution of 3x = 8 is irrational. 3x = 8

Invalid argument: )

log 3x = log 8

) x log 3 = log 8 log 8 ) x= log 3 ) x is irrational.

where both log 8 and log 3 are irrational.

The last step is not valid. The argument that an irrational divided by an irrational is always p p2

irrational is not correct. For example, Dividing by zero is not a valid operation.

2

= 1, and 1 is rational.

a is not defined for any a 2 R , in particular 00 = 6 1. 0

Example 8: Invalid “proof” that 5 = 2 0=0 )

0£5=0£2

)

0£5 0£2 = 0 0

)

fdividing through by 0g

5 = 2, which is clearly false.

This invalid step is not always obvious, as illustrated in the following example. Example 9: Invalid “proof” that 0 = 1: Suppose a = 1 )

a2 = a

)

a2 ¡ 1 = a ¡ 1

) (a + 1)(a ¡ 1) = a ¡ 1 )

a + 1 = 1 .... (¤)

)

a=0 So, 0 = 1

The invalid step in the argument is (¤) where we divide both sides by a ¡ 1. Since a = 1, a ¡ 1 = 0, and so we are dividing both sides by zero.

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Another trap to be avoided is to begin by assuming the result we wish to prove is true. This readily leads to invalid circular arguments.

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APPENDICES

Example 10: Prove without decimalisation that

p 3¡1 >

Invalid argument: p 3 ¡ 1 > p12 ³ ´2 p ) ( 3 ¡ 1)2 > p12 p ) 4 ¡ 2 3 > 12 p 7 ) 2 >2 3 p ) 7>4 3

p1

2

.

fboth sides are > 0, so we can square themg

)

72 > 48

fsquaring againg

)

49 > 48

which is true.

p Hence 3 ¡ 1 >

p Although 3 ¡ 1 >

p1

2

p1

135

is true.

2

is in fact true, the above argument is invalid because we began by

assuming the result. A valid method of proof for

p 3¡1>

p1

2

can be found by either:

² reversing the steps of the above argument, or by p ² using proof by contradiction (supposing 3 ¡ 1 6

p1

2

).

It is important to distinguish errors in proof from a false conjecture. Consider the table alongside, which shows values of n2 ¡ n + 41 for various values of n 2 N . From the many examples given, one might conjecture: “For all natural numbers n, n2 ¡ n + 41 is prime.” This conjecture is in fact false. For example, for n = 41, n2 ¡ n + 41 = 412 is clearly not prime.

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It takes only one counter-example to prove a conjecture is false.

n 1 2 3 4 5 6 7 8 9 10 11 12 13 ¢¢¢ 30 ¢¢¢ 99 ¢¢¢

n2 ¡ n + 41 41 43 47 53 61 71 83 97 113 131 151 173 197 ¢¢¢ 911 ¢¢¢ 9743 ¢¢¢

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136

APPENDICES

IMPLICATIONS AND THEIR CONVERSE If .... then .... Many statements in mathematics take the form of an implication “If A then B”, where A and B are themselves statements. The statement A is known as the hypothesis. The statement B is known as the conclusion. Implications can be written in many forms in addition to “If A then B”. For example, the following all have the same meaning:

8 9 implies > > > > > < so > = hence A B. > > > > thus > > : ; therefore Given a statement of the form “If A then B”, we can write a converse statement “If B then A”. If we know the truth, or otherwise, of a given statement, we can say nothing about the truth of the converse. It could be true or false. A statement and its converse are said to be (logically) independent. For example, suppose x is an integer. ² The statement odd” is false. ² The statement even” is true. ² The statement is also true. ² The statement is also false.

“If x is odd, then 2x is even” is true, but its converse “If 2x is even, then x is “If 2x is even, then x is odd” is false, but its converse “If x is odd, then 2x is “If x > 1, then ln x > 0” is true, and its converse “If ln x > 0, then x > 1” “If x = 5, then x2 = 16” is false, and its converse “If x2 = 16, then x = 5”

EXERCISE Prove or disprove: 1 If x is rational then 2x 6= 3. 2 If 2x 6= 3 then x is rational.

EQUIVALENCE Some conjectures with two statements A and B involve logical equivalence or simply equivalence. We say A is equivalent to B, or A is true if and only if B is true. The phrase “if and only if” is often written as “iff” or ,.

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A , B means A ) B and B ) A

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APPENDICES

In order to prove an equivalence, we need to prove both implications: A ) B

137

and B ) A.

For example: x2 = 9 , x = 3 is a false statement. x = 3 ) x2 = 9 is true but x2 = 9 6) x = 3 as x may be ¡3. Example 11: Prove that (n + 2)2 ¡ n2 is a multiple of 8 , n is odd. Proof: ())

(n + 2)2 ¡ n2

is a multiple of 8

2

) n + 4n + 4 ¡ n2 = 8a for some integer a ) 4n + 4 = 8a ) n + 1 = 2a ) n = 2a ¡ 1 ) n is odd. (()

n is odd ) n = 2a ¡ 1 for some integer a ) n + 1 = 2a ) 4n + 4 = 8a ) (n2 + 4n + 4) ¡ n2 = 8a ) (n + 2)2 ¡ n2

is a multiple of 8.

In the above example the ()) argument is clearly reversible to give the (() argument. However, this is not always the case. Example 12: Prove that for all x 2 Z + , x is not divisible by 3 , x2 ¡ 1 is divisible by 3. Proof: ())

x is not divisible by 3 ) either x = 3k + 1 or x = 3k + 2 for some k 2 Z + [ f0g ) x2 ¡ 1 = 9k2 + 6k or 9k2 + 12k + 3 = 3(3k2 + 2) or 3(3k 2 + 4k + 1) ) x2 ¡ 1 is divisible by 3.

(()

x2 ¡ 1 is divisible by 3 ) 3 j x2 ¡ 1 ) 3 j (x + 1)(x ¡ 1) ) 3 j (x + 1) or 3 j (x ¡ 1) fas 3 is a prime numberg ) 3-x

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or in other words, x is not divisible by 3.

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APPENDICES

NEGATION For any given statement A, we write not A or :A to represent the negation of the statement A. A x>0

:A x60

x is prime

x is not prime

x is an integer

x is not an integer

x is rational z is real

x is irrational z = a + bi, a, b 2 R , b 6= 0

x is a multiple of 3

x is not a multiple of 3 or x = 3k + 1 or 3k + 2 for k 2 Z + [ f0g

For example:

For x 2 R : For z 2 C : For x 2 Z + [ f0g:

PROOF OF THE CONTRAPOSITIVE To prove the statement “If A then B”, we can provide a direct proof, or we can prove the logically equivalent contrapositive statement “If not B, then not A” which we can also write as “If :B, then :A”. For example, the statement “If it is Jon’s bicycle, then it is blue” is logically equivalent to “If that bicycle is not blue, then it is not Jon’s”. Example 13: Prove that for a, b 2 R , “ab is irrational ) either a or b is irrational”. Proof using contrapositive: a and b are both rational ) a =

p r and b = where p, q, r, s 2 Z , q 6= 0, s 6= 0 q s

) ab =

µ ¶³ ´ p q

r s

=

pr fwhere qs 6= 0, since q, s 6= 0g qs

) ab is rational

fsince pr, qs 2 Z g

Thus ab is irrational ) either a or b is irrational. Example 14: Prove that if n is a positive integer of the form 3k + 2, k > 0, k 2 Z , then n is not a square. Proof using contrapositive: If n is a square then n has one of the forms (3a)2 , (3a + 1)2 or (3a + 2)2 , where a 2 Z + [ f0g. ) n = 9a2 , 9a2 + 6a + 1 or 9a2 + 12a + 4 ) n = 3(3a2 ), 3(3a2 + 2a) + 1 or 3(3a2 + 4a + 1) + 1 ) n has the form 3k or 3k + 1 only, where k 2 Z + [ f0g ) n does not have form 3k + 2.

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Thus if n is a positive integer of the form 3k + 2, k > 0, k 2 Z , then n is not a square.

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APPENDICES

139

USING PREVIOUS RESULTS In mathematics we build up collections of important and useful results, each depending on previously proven statements. Example 15: Prove the conjecture: “The recurring decimal 0:9 = 0:999 999 99:::: is exactly equal to 1”. Proof (by contradiction): Suppose 0:9 < 1 ) 0:9
1 is absurd, 0:9 = 1. Proof (Direct Proof): 0:9 = 0:999 999 99:::: = 0:9 + 0:09 + 0:009 + 0:0009 + :::: ¡ ¢ 1 1 1 = 0:9 1 + 10 + 100 + 1000 + :::: µ1 ¶ P ¡ 1 ¢i 9 = 10 10 i=0 Ã !

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1 1 ¡ 10

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APPENDICES

THEORY OF KNOWLEDGE

AXIOMS AND OCCAM’S RAZOR

In order to understand complicated concepts, we often try to break them down into simpler components. But when mathematicians try to understand the foundations of a particular branch of the subject, they consider the question “What is the minimal set of assumptions from which all other results can be deduced or proved?” The assumptions they make are called axioms. Whether the axioms accurately reflect properties observed in the physical world is less important to pure mathematicians than the theory which can be developed and deduced from the axioms. Occam’s razor is a principle of economy that among competing hypotheses, the one that makes the fewest assumptions should be selected. 1 What value does Occam’s razor have in understanding the long-held belief that the world was flat? 2 Is the simplest explanation to something always true? 3 Is it reasonable to construct a set of mathematical axioms under Occam’s razor? One of the most famous examples of a set of axioms is given by Euclid in his set of 13 books called Elements. He gives five axioms, which he calls “postulates”, as the basis for his study of Geometry: 1. Any two points can be joined by a straight line. 2. Any straight line segment can be extended indefinitely in a straight line. 3. Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as centre. 4. All right angles are congruent. 5. Parallel postulate: If two lines intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough. 4 Is the parallel postulate genuinely an axiom, or can it be proved from the others? 5 What happens if you change the list of axioms or do not include the parallel postulate?

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6 What other areas of mathematics can we reduce to a concise list of axioms?

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APPENDICES

APPENDIX B:

FORMAL DEFINITION OF A LIMIT

The informal definition of a limit we were given in Section B is sufficient for our study here. However, for completeness we justify its use by considering and interpreting the formal definition: The formal definition of a limit of a function: Let f be a function defined in an open interval about x = a, except f (a) need not be defined. The number l is called the limit of f as x approaches a if, for any value " > 0, there exists a corresponding value ± > 0 such that 0 < jx ¡ aj < ± ensures jf(x) ¡ lj < ". We write lim f(x) = l to denote “the limit of f as x approaches a, is l”. x!a

To interpret this definition, note that l and a are constants, and x and f(x) are variables. The value " > 0 is any positive value, but in particular we consider " a very, very small positive value. In this case

y

jf(x) ¡ lj < " , ¡" < f (x) ¡ l < " , l ¡ " < f (x) < l + " , the value f (x) is very, very close to l

l+" l l-"

f (x) will lie in the shaded band of values shown.

x

Similarly, the value ± > 0 is a positive value, and in particular we consider ± a very, very small positive value. y 0 < jx ¡ aj < ± , x 6= a and ¡± < x ¡ a < ± , x 6= a and a ¡ ± < x < a + ± so that x is very, very close to, but not equal to, the value a.

In this case

a-± a a+±

x

Putting this all together, we have that f has limit l as x approaches a if for any choice of " > 0, there always exists a value ± > 0 such that the value f (x) will be within distance " from l, whenever x is within distance ± from a, but not equal to a. In particular, this must be true for " > 0 chosen as small as we like. y y = f(x) l+" l l-"

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a-± a a+±

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142

APPENDICES

These algebraic and geometric interpretations of the formal definition lead to the informal definition of the limit of a function given in Section B.

ESTABLISHING LIMITS By considering the identity function f (x) = x, we intuitively see that lim x = a for any constant a 2 R .

y

x!a

This is proved formally as follows: y = f(x) = x

l=a

Suppose f(x) = x and a 2 R is a constant. For any " > 0, there exists ± = " > 0 such that 0 < jx ¡ aj < ± ) jx ¡ aj < "

fsince ± = "g

) jf (x) ¡ aj < "

fsince f (x) = xg

a

x

Since an appropriate ± can be found for any choice of " > 0, by the formal definition of a limit, lim x = a, for a any constant.

x!a

Arguments such as this can be used to establish the limit laws presented in Section B. When we define new functions using a sum, composition, or some other combination of simpler functions, the limit laws help us calculate limits for these new functions using what we already know about the simpler functions. 3(x2 ¡ 1) . x!1 x¡1

Example: Determine the limit lim Using Limit Laws 3(x2 ¡ 1) x¡1 3(x ¡ 1)(x + 1) = lim x!1 (x ¡ 1)

lim

x!1

= lim 3(x + 1)

fusing Limit Laws with

x!1

x¡1 = 1 a constant, since x 6= 1g x¡1

= lim (3x + 3) x!1 ³ ´³ ´ ³ ´ 3 lim x + lim 3 fby the Limit Laws, since each limit existsg = xlim !1 x!1 x!1 fsince lim 3 = 3 from the Limit Laws

=3£1+3

x!1

and lim x = 1 from above using a = 1g

=6

x!1

Using the formal definition directly (not required for this course) From the graph of y =

3(x2 ¡ 1) we observe that for x near 1, f (x) is close to the value 6. x¡1

3(x2 ¡ 1) = 6, and must prove this formally. x!1 x¡1

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We conjecture that lim

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143

¯ ¯ ¯ 3(x2 ¡ 1) ¯ ¯ For x 6= 1, jf (x) ¡ 6j < " , ¯ ¡ 6¯¯ < " x¡1 ¯ ¯ ¯ 3(x ¡ 1)(x + 1) ¯ ¯ ¡ 6¯¯ < " , ¯ (x ¡ 1) , , , ,

j3(x + 1) ¡ 6j < " j3x + 3 ¡ 6j < " j3x ¡ 3j < " 3 jx ¡ 1j < "

, jx ¡ 1j
0 such that 0 < jx ¡ 1j < ± implies jf (x) ¡ 6j < ". 3 3(x2 ¡ 1) = 6. Thus by the formal definition of the limit, lim x!1 x¡1

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Therefore, for all " > 0, there exists ± =

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WORKED SOLUTIONS 6 jx ¡ aj
0, jaj = a ) jaj > 0 .... (1) If a < 0, jaj = ¡a where ¡a > 0 ) jaj > 0 .... (2) From (1) and (2), jaj > 0 for all a 2 R . if ¡a > 0 0 if ¡a = 0 ¡(¡a) if ¡a < 0

= =

n" > 1 ) " >

if a < 0 if a = 0 if a > 0

0 a

na

= jaj

(2) If Pk is true, then (1 + x)k > 1 + kx, k 2 Z + .

or j¡aj = j¡1 £ aj = j¡1j jaj = 1 jaj = jaj

Now x > ¡1 ) 1 + x > 0

f jabj = jaj jbj g

)

)

y

(1 + x)k+1 > 1 + (k + 1)x, as kx2 > 0

® lies in A, but is smaller than ® 2

) ) )

r + x is irrational. c Similarly suppose rx = , c, d 2 Z , d 6= 0 d ) x=

a

y

x

b

1

a

)

lim (¡x) DNE.

x!1

x y = -x

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As x ! 1, ¡x ! ¡1

y

fTriangle inequalityg

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c which 2 Q , a contradiction dr

) rx is irrational.

If two points lie in a particular interval [a, b] on the number line then the distance between them is less than the length of the interval.

0

a where a, b 2 Z , b 6= 0 b a x = ¡ r which 2 Q , a contradiction b

r+x =

EXERCISE B.1

b

5 ja ¡ bj = j(a ¡ c) + (c ¡ b)j 6 ja ¡ cj + jc ¡ bj

® _ "w

¼¡ , y ! 1 below 2

lim f (x) = 1 ¼¡

As x !

x

´ 1

lim f (x) = 1 ¼+

x! 2

c Since

lim f (x) = ¼¡

x! 2

lim f (x) = 1, ¼+

x! 2

lim f (x) = 1 ¼

x! 2

DNE.

x

¼+ , y ! 1 below 2

4

a

y

y = -Qv

2

a

x

x! 2

) d

2¼

x+1 y = ^^^% x - 1_ y=1

y -1 -1

y = 3x + 2

x

x

(-1, -1)

x=1

b ¡

As x ! ¡1 , As x ! ¡1+ , )

ii As x ! 1, y ! 1 above iii As x ! 1¡ , As x ! 1+ ,

lim (3x + 2) = ¡1

x!¡1

b f (x) = )

3x + 2 ! ¡1 below 3x + 2 ! ¡1 above

(x + 2)(x ¡ 1) x2 + x ¡ 2 = x+2 x+2

f (x) =

i As x ! ¡1, y ! 1 below

5

y ! ¡1 y!1

x2 + x - 2_ f(x) = ^^^^^^%^ x+2

(3, 5)

x

y = 3x, x < 3 3

y ! 9 below y ! 9 above

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c Yes the limit exists fas both LH and RH limits are 9g lim f (x) = 9

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y ! ¡3 below y ! ¡3 above

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As x ! ¡2¡ , As x ! ¡2+ ,

lim f (x) DNE.

x!1

(3, 9) missing point

y

-1

)

lim f (x) = 1

x!1

y = x2, x > 3

if x 2 R , x 6= ¡2 undefined if x = ¡2

missing point (-2, -3)

)

lim f (x) = 1

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1

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WORKED SOLUTIONS p x does not exist

6 For x < 0, )

lim

lim

x+1 x2 ¡ 2x ¡ 3 x+1 = lim x!¡1 (x + 1)(x ¡ 3) lim

x!¡1

DNE.

x!0¡

So, although 7

b

p

x!0+

x = 0, lim

p

x DNE.

x!0

a

= ¡ 14

y y = 1, x > 0

1

=

-1

d

=5

lim f (x) = ¡1

x!0¡

lim f (x) = 1

=

x!0+

=

iii Since the limits in i and ii are different, lim f (x) DNE. x!0

b

=

y

¡

1 1 ¡ 15 ¡3 2 3 ¡ 13 £ 10 1 ¡ 10

y!5

1 -"w

¢

µ

1 1 ¡ y 5

1 1 ¡ y 5

5¡y 5y(y ¡ 5)1 ¡1 = lim y!5 5y

x2 - 3x + 5 f(x) = ^^^%^^%^ , x > 0 5__ _ _ _ _ _ _ _ _ _ _ _ _ _ _ x

-2¼

µ

1 y¡5

lim

f

f(x) = sin_x, x < 0

-¼

1 lim y!2 y ¡ 5

e

ii As x ! 0+ , y = 1

-Es"

x2 + 3x ¡ 4 x¡1 (x ¡ 1)(x + 4) = lim x!1 x¡1 lim

x!1

i As x ! 0¡ , y ! ¡1 below

)

x!0 ¡4 ¡1

=4

x y = (x - 1)3, x < 0

fx 6= ¡1g

x2 + 3x ¡ 4 x¡1

lim

c

)

147

-1

= lim

¶

¶

fy 6= 5g

y!5

-1

fx 6= 1g

1 = ¡ 25

i As x ! 0¡ , y ! 0 ii As x !

0+ ,

)

y!1

)

lim f (x) = 0

x!0¡

lim f (x) = 1

2

x!0+

)

iii Since the limits in i and ii are different, lim f (x) DNE.

)

x!0

c

y

i As x ! 0¡ , y ! 0

)

y!0

)

lim f (x) = 0

=0

x!0¡

d

lim f (x) = 0

x!0+

)

cyan

lim

µ!0¡

DNE

tan x sec x

lim

sin x cos x cos x 1

¼¡ x! 2

=

cos µ µ

¼¡

lim µ = 0

µ!0¡

lim

x! 2

f cos x 6= 0g

magenta

yellow

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

=1

25

0

x+1 x2 ¡ 2x ¡ 3

5

95

50

75

25

0

5

=

100

=

lim cos µ = 1 but

µ!0¡

e

lim

DNE

sin x 0 = =0 ex 1 sin x c lim x!¼¡ 1 ¡ cos x 0 = 1 ¡ (¡1)

EXERCISE B.2 x!1 2 ¡4 ¡ 12

p 2¡x =0

x!0

x!0

a

x!2¡

lim

x!2¡

b lim

iii lim f (x) = 0

1

x!2¡

ln x lim p 2¡x

)

x

ii As x !

lim ln x 6= 0 but

Now f(x) = sin(_Qv_), x 6 = 0

0+ ,

x0 p 2 ¡ x exists.

a

black

Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\147IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 10:45:02 AM BRIAN


148 3

WORKED SOLUTIONS 3+x x+5

lim

a

x!1

3 x

= lim

lim

b

x!1

+1

= lim

5 1+ x

x!1

=1

r

lim

c

x!1

r

= lim

lim

d

x!1 x2

)

= lim

x2 + x ¡ x

= lim

³p

x!1

= lim

x!1

= lim

x!1

x2 + x + x

cos x sin x ³ x ´ = lim cos x x!0 sin x = lim x x!0

=1£1 =1

!

µ

x!0

x!0

= lim

fas x > 0g

x!0

=

+1

g

1 2

lim

x!0+

x!a

, ,

x!a

³ sin x ´ p

x!a

x + sin x = lim a lim x!0 x ¡ sin x x!0

2

But as x ! 0, 1 ¡ )

sin 3µ b lim µ!0 µ ³ sin 3µ ´ = lim 3 µ!0 3µ ³ sin 3µ ´ = 3 lim 3µ!0 3µ

x

x

b lim

h!0

= lim

h!0

´

¡ sin2 h h(cos h + 1)

³ sin h ´2 h

h!0

= 0 £ 12 £

1 cos h + 1

1 2

yellow

95

100

50

75

25

0

5

95

100

50

75

=0

25

0

5

95

100

50

75

25

0

5

100

´³

= lim ¡h

=3

magenta

³

cos h ¡ 1 cos h ¡ 1 cos h + 1 = lim h!0 h h cos h + 1 cos2 h ¡ 1 = lim h!0 h(cos h + 1)

= 3£1

cyan

sin x x sin x 1¡ x 1+

sin x !1¡1=0 x

the limit DNE.

3µ ! 0 alsog

95

sin 2x

£1

x!a

lim (f (x) ¡ l) = 0

fas µ ! 0,

50

sin 2x 1 2

³ 2x í

1 2

x!a

sin2 µ a lim µ!0 µ ³ sin µ ´ = lim sin µ µ!0 µ

75

2

+

lim f (x) ¡ lim l = 0

=1£0 =0

25

h x ³ 2x ´

¶

lim f (x) = lim l

x!a

EXERCISE B.3

0

x2 x + sin 2x sin 2x

=1£0 =0

5 If lim f (x) = l ,

5

¶

sin x p = lim x x!0+

1 2

=

µ

x2 + x sin 2x

=0£1+

1 = 1+1

1

lim

f

1

1+

£1

lim x cot x

= lim

1 1+ x +x 1 x

=1

=

7 4 7 4

x!0

x2 + x + x

q q

=

e

p

x

=1£1

µ!0

p

x2 + x + x x

cos µ

fSince lim

x2 + x ¡ x2

= lim

x!1

´

x2 + x ¡ x

sin µ

sin µ = 1, µ µ = 1g lim µ!0 sin µ

1+x +x+1

0+0 1+0+0

Ãp

sin 7x 4x ³ sin 7x ´ 7 = lim 4 x!0 7x ³ sin 7x ´ = 74 £ lim 7x!0 7x lim

x!0

³ µ ´

µ!0

p

x!1

d

cos µ

=0

the limit DNE.

lim

4

=

µ tan µ µ

µ!0 sin µ

= lim

3 !1 x

x+

= lim

1 1 +x x2 1 x!1 1+ x + x12

3 x

But as x ! 1,

q

5 4¡ x + x12

1 1+ x + x12 4¡0+0 = 1+0+0

+3 x

x+

x!1

µ!0

=4

x2

lim

c

x!1

0+1 1+0

=

4x2 ¡ 5x + 1 x2 + x + 1

black

Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\148IB_HL_OPT-Calculus_an.cdr Thursday, 7 March 2013 9:04:48 AM BRIAN


WORKED SOLUTIONS

³

´³

1 ¡ cos x 1 ¡ cos x 1 + cos x = lim x!0 x2 x2 1 + cos x sin2 x = lim 2 x!0 x (1 + cos x)

c lim

x!0

³ sin x ´2

= lim 1

x

x!0

= 1 £ 12 £

´

Now 0 6 jxj 6 2x )

0 jxj 2x 6 6 1 + x4 1 + x4 1 + x4 0 2x ! =0 1 + x4 1

As x ! 0+ ,

1 1 + cos x

)

1 2

1 2

=

149

lim

x!0+

jxj = 0 .... (1) fSqueeze Theoremg 1 + x4

Consider x < 0

y = -2x

y y = |x|

3

³1´

a As ¡1 6 cos

y=0

³1´

¡x2 6 x2 cos

x

6 1, then

x2

x2

6 x2

But lim (¡x2 ) = lim (x2 ) = 0 x!0

³1´

lim x2 cos

)

Now 0 6 jxj 6 ¡2x

x!0

x2

x!0

³1´

b As ¡1 6 sin

³1´ x

6 x fif x > 0g

)

But lim (¡x) = lim (x) = 0 x!0

)

lim x sin

x!0

However, if x < 0 ³ ´ ³ ´ 1 1 ¡1 6 sin 6 1 ) x 6 x sin 6 ¡x x x

EXERCISE C.1 1

a Let F (x) = f (x)g(x) As f (x) and g(x) are defined at x = a, F (a) = f (a)g(a) is also defined at x = a.

and as lim (¡x) = lim (x) = 0, x!0

)

x!0

³1´

lim x sin

x

x!0

jxj = 0 .... (2) fSqueeze Theoremg 1 + x4

= 0 fSqueeze Theoremg

x

x!0

lim

x!0¡

Now lim F (x) = lim [f (x)g(x)]


x!a

x!a

= lim f (x) £ lim g(x)

1

x!a

c As ¡1 6 sin x 6 1 and e x > 0 for all x 2 R , ¡

1 1 ex

sin x

6

e

6

1 x

1

= F (a)

1 ex

) 1

Ã )

lim

¡

x!0

¡ )

lim e

!

1 e

1 ¡x

Ã

= lim

1 x

x!0

¢

x!0

1 e

As f (x) and g(x) are defined at x = a, F (a) = f (a) § g(a) is also defined at x = a.

=0

1 x

f (x) g(x) is continuous at x = a.

b Let F (x) = f (x) § g(x)

!

Now lim F (x) = lim (f (x) § g(x)) x!a

x!a

= lim f (x) § lim g(x)

sin x = 0 fSqueeze Theoremg

x!a

y = 2x

y

= F (a) )

y = |x| y=0

f (x) § g(x) is continuous at x = a.

c Let F (x) =

x

f (x) where g(x) 6= 0 g(x)

Now F (a) =

magenta

yellow

95

f (a) where g(a) 6= 0 g(a)

F (a) is defined.

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

)

5

x!a

= f (a) § g(a)

d Consider x > 0

cyan

x!a

= f (a) £ g(a)

1 ! 1 and e x ! 1 x

But as x ! 0,

¡2x !0 1 + x4

From (1) and (2), lim g(x) = 0

x!0

³1´

0 jxj ¡2x 6 6 1 + x4 1 + x4 1 + x4

But, as x ! 0¡ ,

6 1, then

x

¡x 6 x sin

)


black

Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\149IB_HL_OPT-Calculus_an.cdr Thursday, 7 March 2013 1:44:34 PM BRIAN


150

WORKED SOLUTIONS Now lim F (x) = lim x!a

x!a

f (x) g(x)

4

y

a 12

y = x2

lim f (x)

x!a

=

(3, 9)

9

lim g(x)

x!a

)

6

f (a) g(a)

=

3

= F (a) f (x) is continuous at x = a. g(x)

y = 3x

½

Now lim F (x) = lim c f (x)

f (x) =

x!a

= c lim f (x) x!a

b

= c f (a)

y=5

5

e Let F (x) = [f (x)]n , n 2 Z +

2

) F (a) = [f (a)]n is defined as f (a) is defined. Now lim F (x) = lim [f (x)]n

h

x!a

=

lim f (x)

[f (x)]n

3

-15

f (x) is continuous for all x 2 R , x 6= 3. At x = 3 we have a ‘jump’ discontinuity which is essential.

a

a

y

=1+1+1 =3 ) let k = 3

5__ _ _ _ _ y = ^^^^ 4-x

b

sin 3x x ³ sin 3x ´ = lim 3 x!0 3x ³ sin 3x ´ = 3 £ lim 3x!0 3x lim

x!0

y = 2x

8 y=5

4

x3 ¡ 1 x¡1 (x ¡ 1)(x2 + x + 1) = lim x!1 x¡1 lim

x!1

i f has an essential discontinuity at x = 6. ii f has a removable discontinuity at x = 5. b f is continuous on fx j x 2 R , x 6= 5 or 6g

=3£1

c f (2) = 22 = 4 ) let k = 4 1 d y = has an essential discontinuity at x = 0 which x cannot be removed ) no value for k can be found.

-4

x=4

cyan

magenta

yellow

50

25

0

e y = kx represents the infinite number of lines passing through O(0, 0). ) k2R

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

f is continuous at x = ¡2. f has a ‘jump’ discontinuity at x = 0. f has a removable discontinuity at x = 3. f has a ‘break’ discontinuity at x = 4 due to the vertical asymptote at x = 4.

100

50

75

25

0

5

i ii iii iv

fas x ! 0, 3x ! 0g

=3 ) let k = 3

x

95

4

100

2

75

-2

y = x2 - 6

b

x

x!a

= F (a) is continuous at x = a.

y=x

8

(3, -14)

a

-4

6

y = x2 - 10x + 7

-10

in

5 2

4

-5

= [f (a)]n )

x>3 x 0, then k > ¡2, and f (x) has graph: y

) x

1 n

1

= (eln x ) n

1

1

) xn = en

(2, -k - 2)

1

x

1

y=x+k (2, k + 2)

1

lim f (g(n)) = f

From 7 c,

)

a f (x) is continuous for all x 2 R and ) on [0, 2]. f (0) = ¡3 and f (2) = 7 Thus, with k = 0, f (0) 6 k 6 f (2) Hence, by the IVT, there exists c 2 [0, 2] such that f (c) = 0 ) f (x) has a zero on [0, 2]. y b

1 Z+

But g(a) = 1 since a 2 Q lim g(xn ) 6= g(a)

n!1

lim xn = a

n!1

)

(3, 4)

g(x) is discontinuous at x = a.

b If xn is the decimal expansion of a to n decimal places then xn 2 Q )

1

x

(1, -4)

lim g(xn ) = 1 fDirichlet functiong

x=2

f (x) has an essential discontinuity at x = 2 ) the IVT will not apply. Clearly for [1, 2[ , f (x) < 0 and for ]2, 3], f (x) > 0 ) f (x) is never zero on [1, 3].

lim g(xn ) 6= g(a)

n!1

lim xn = a

Also

n!1

)

3

n!1

But g(a) = 0 since a 2 =Q )

= f (ln 1)

EXERCISE C.2

lim g(xn ) = 0 fDirichlet functiong

But

n!1

n!1

= e0 =1

n!1

)

lim x

1 n

´

lim g(n)

= f (0)

p 2 , a2Q n

Since a 2 Q , xn 2 = Q for all n 2 )

³

n!1

) f (x) is discontinuous f (x) is continuous at x = 2 for all x 2 R So, if k > ¡2, k 2 R , f (x) is continuous and if k < ¡2, k 2 R , f (x) is continuous for all x 6= 2. )

a xn = a +

1 n

b From a, x n = f (g(n)) where f (n) = en , g(n) = ln x n

x

6

ln x

) x n = eln x

y = -k - 2

(2, k + 2)

x = eln x

a

y y=x+k

y=k+2

8

151

g(x) is discontinuous at x = a.

2 f (x) = x3 ¡ 9x2 + 24x ¡ 10 has graph:

c From a and b, the Dirichlet function is nowhere continuous.

y

7

(2, 10)

(5, 10)

(1, 6)

(4, 6)

a If g(x) is continuous at x = a, then lim g(x) = g(a) = l

.... (1)

x!a

If f (x) is continuous at g(a) then x

f (g(x)) = f (g(a)) = f (l)

f 0 (x) = 3x2 ¡ 18x + 24 ) (2, 10) is a local maximum (4, 6) is a local minimum. = 3(x2 ¡ 6x + 8) = 3(x ¡ 2)(x ¡ 4) a On [1, 5], m = 10 and xm = 2 or 5 + - + b On [1, 5], m = 6 and xm = 1 or 4 2 4

But g(x) ! g(a) when x ! a lim f (g(x)) = f (l)

x!a

b From a, if lim g(x) exists then, x!a

x!a

cyan

´

lim g(x)

magenta

yellow

25

0

5

95

x!1

100

95

50

75

25

0

5

95

100

50

75

25

0

5

x!1

³

50

lim f (g(x)) = f (l) = f

75

x!1

25

lim g(x) exists then

3 As f (a) and f (b) have opposite signs, then f (a) 6 f (b) or f (b) 6 f (a) By the IVT, there exists c 2 [a, b] such that f (a) 6 f (c) = 0 6 f (b) or f (b) 6 f (c) = 0 6 f (a) ) there exists a zero c 2 [a, b] ) there exists at least one zero of f (x).

ffrom (1)g

95

´

lim g(x)

0

c If

³

5

=f

100

x!a

50

lim f (g(x)) = f (l)

100

)

75

lim

g(x)!g(a)

black



152

WORKED SOLUTIONS a Let f (x) = sin x + rx ¡ 1, r > 0 for x 2 ]0,

4

Now f (0) = ¡1 < 0 and f Since f (0) and f exists c 2 [0,

³¼´ 2

³¼´ 2

n

¼ [. 2

h!0

¼ =r >0 2

2 f (5) = 0 is defined. lim f (x) = lim (5 ¡ x) = 0 and

x!5¡

are opposite in sign, by the IVT there

y y = sin_x 1

A

-2 2

1

lim f (x) = lim (x ¡ 5) = 0

)

a If f (x) =

5

x!5

Thus f (x) is continuous at x = 5.

) f (x) = 0 ) x ¼ 0:511

Now f (x) =

) f (x) is not differentiable at x = 5 3 f (x) =

4 f (x) =

= ¡1 + )

y

(1, 13)

(1, 10) y = 3x + 10, x1

¡1 6 f (¡1) 6 ¡0:4146

0 (1) = 3 i f¡

b

Since f is continuous for all x, r 2 R , f is also continuous on [¡1, 1] and since f (¡1) < 0 < f (1) )

x>1 x0

0 (0) 6= f 0 (0) f+ ¡

1 + 1 + sin2 (1) + (r + 1)2

and also

n x + 2,

x2 + 3x, x < 0 has an essential ‘jump’ discontinuity at x = 0 ) f (x) is not continuous at x = 0 ) f (x) is not differentiable at x = 0 0 (0) = 0 and f 0 (0) = 2(0) + 3 = 3 and Note: f+ ¡

for x = §1 and r 2 R . f (1) = (1)

x>5 x 0 for r 2 R 1 ) 06 6 0:5854 1 + sin2 x + (r + 1)2

15

n x ¡ 5,

0 0 ) f¡ (x) = ¡1 and f+ (x) = 1

x

1 + 1 + sin2 x + (r + 1)2

So, for r 2 R ,

x!5+

lim f (x) = 0 = f (5)

When r = 1, at A, sin x = 1 ¡ x

y=1-x

x15

x!5¡

x!5+

¼ ] such that f (c) = 0 2

b

o

sin h cos h ¡ 1 = 1 and lim = 0 fExample 6g h h!0 h

lim

0 (1) = ¡2(1) + 5 = 3 ii f+ 0 (1) = f 0 (1), f (x) is not continuous at c No, although f¡ + x = 1.

there exists an x 2 [¡1, 1] such that f (x) = 0.

1 , where r 2 R , r > 0, there is a 5x discontinuity at x = 0. f is not continuous on the interval [¡1, 1], so the IVT cannot be used to prove the existence of a zero.

b If g(x) = rx17 +

5

a

y 2 y = x2 + x + 1,

EXERCISE D 1 If f (x) = cos x

y = 1 + sin_x, x > 0

1

x 0

(0, 1)

x!1¡

lim f (x) = lim (cx + d) = c + d

and

) f (x) is differentiable at x = 0 b

lim f (x) = lim x2 = 1

Also,

0 0 ) f¡ (0) = f+ (0) = 1

) and

= 8(2) = 16

x!0

0 6= f+ (2)

x!0

) lim f (x) = 0

fSqueeze Theoremg

x!0

) f (x) is not differentiable at x = 2

) lim f (x) = f (0) x!0

) f (x) is continuous at x = 0

6 f (0) = k sin 0 = 0 ) f (0) is defined. Also,

f 0 (x) = 3x2 sin

lim f (x) = lim (k sin x) = k(0) = 0

where

x!0¡

and

x!0¡

x!0+

)

Consider x > 0

lim f (x) = lim (tan x) = tan 0 = 0 x!0+

¡1 6 sin

lim f (x) = 0 = f (0)

x!0

) f (x) is continuous at x = 0 for all k 2 R .

and

1

¡1 6 cos

0 Now f¡ (0) = sec2 (0) = 2 = 1 and 1

³1´ x

³1´ x

³ ´ 1 x

6 1 ) x > ¡x cos

7

Thus,

f (x) is differentiable at x = 0 , k = 1

a f (x) =

x2 , cx + d,

¡3x2 ¡ x 6 3x2 sin

But

x

¡ x cos

³1´

95

x

> ¡x

x ³ ´ 1 x

³1´ x

6 3x2

6x

6 3x2 + x

x!0+

0

lim f (x) = 0 .... (1) fSqueeze Theoremg

x!0+

100

50

75

25

0

5

95

50

75

25

0

100

yellow

³1´

lim (¡3x2 ¡ x) = 0 = lim (3x2 + x)

x!0+

)

5

95

50

75

25

0

5

95

100

50

75

25

0

5

100

magenta

³1´

or ¡3x2 ¡ x 6 f 0 (x) 6 3x2 + x

x61 x>1

f (1) = 12 = 1 is defined.

cyan

1 x

) ¡x 6 ¡x cos

0 0 ) f¡ (0) = f+ (0) , k = 1

½

³ ´

6 1 ) ¡3x2 6 3x2 sin

0 (0) = k cos(0) = k f+

)

¡ x cos

black



154

WORKED SOLUTIONS )

Consider x < 0 as 3x2 > 0, 2

2

¡3x 6 3x sin and as ¡x > 0,

³1´ x

³1´

x 6 ¡x cos

x

Thus, ¡3x2 + x 6 3x2 sin

³1´ x

2

)

2

6 3x

lim

ex ex = lim which does not exist x x!1 1

lim

ex x

x!1

x!1

e As x ! 0+ , x ! 0, ln x ! ¡1 The limit has type 0 £ 1, so we can use l’Hôpital’s Rule.

6 ¡x

¡ x cos

³1´ x

2

)

lim x ln x = lim

x!0+

x!0+

= lim

x!0+

lim (¡3x2 + x) = 0 = lim (3x2 ¡ x)

x!0¡

=0

x!0¡

f As x ! 0, arctan x ! 0

b

f 0 (x)

=

x!0

3x2 sin

³1´ x

0,

¡ x cos

³1´ x

The limit has type ,

x 6= 0

)

x=0

lim

x!0

x!0+

)

lim f 0 (x) = 0 = f 0 (0)

x!0¡

lim

)

lim

x!0+

a As x ! 0, 1 ¡ cos x ! 0 and x2 ! 0

x!0

= 0 , 0

Note:

x!0

=

The limit has type )

so we again use l’Hôpital’s Rule.

0 , 0

lim

x!1

The limit has type 0 £ 1, so we can use l’Hôpital’s Rule.

1 2

)

lim x2 ln x = lim

x!0+

= lim

x!0+

1 x

¡2x¡3

= lim ¡ 12 x2 x!0+

magenta

yellow

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

so we can use l’Hôpital’s Rule.

100

50

1 , 1

ln x x¡2

=0

75

25

0

5

95

100

50

75

25

0

5

x!0+


d As x ! 1, ex ! 1

cyan


j As x ! 0+ , x2 ! 0 and ln x ! ¡1

ln x 1 x = lim = lim =1 x!1 x x ¡ 1 x!1 1

The limit has type

0 , 0

x + sin x 1 + cos x = lim lim x!0 x ¡ sin x x!0 1 ¡ cos x

1

)

´

which DNE as 1 + cos x ! 2 and 1 ¡ cos x ! 0

c As x ! 1, ln x ! 0 and x ¡ 1 ! 0 The limit has type

x!0+

³

p sin x sin x x p = lim + x x x!0

i As x ! 0, x + sin x ! 0 and x ¡ sin x ! 0

ex ¡ 1 ¡ x ex = lim 2 x!0 2 x

lim

lim

=0£1 =0

ex ¡ 1 ¡ x ex ¡ 1 = lim lim x!0 x!0 x2 2x )


=0

£1

0 , 0

0 , 0

p x!0

x!0+


This limit also has type

1 2

= 2(0)(1)

b As x ! 0, ex ¡ 1 ¡ x ! 1 ¡ 1 ¡ 0 = 0 and x2 ! 0 The limit has type


p = lim 2 x cos x

1 ¡ cos x sin x = lim x!0 2x x2 sin x = 12 lim x!0 x 1 2 1 2

=1

sin x cos x p = lim 1 x x!0+ 1 ¡2 x 2


=

¶

1

h As x ! 0+ , sin x ! 0 and

EXERCISE E

lim

1 1+x2

+x 2x + 1 = lim x!0 2 cos 2x sin 2x

x!0

The limit has type

)

0 , 0

=

) f 0 (x) is continuous for all x

0 , 0

µ

x2

lim f 0 (x) = f 0 (0)

x!0

The limit has type


arctan x = lim x x!0

The limit has type

lim f 0 (x) = 0 = f 0 (0) and

1

0 , 0

g As x ! 0, x2 + x ! 0 and sin 2x ! 0

c From (1) and (2) above,

)

¡x¡2

x!0+

x!0¡

So, for x 6= 0, lim f 0 (x) = 0

(

1 x

= lim (¡x)

lim f 0 (x) = 0 .... (2) fSqueeze Theoremg

)

ln x x¡1

6 3x2 ¡ x

or ¡3x + x 6 f (x) 6 3x ¡ x But

DNE

black



155

WORKED SOLUTIONS k As x ! 0, ax ¡ bx ! 0 and sin x ! 0 0 , 0


ax

lim

¡ bx

The limit has type

ax ln a ¡ bx ln b

= lim

sin x

x!0

a As x ! 0+ , ln(cos 5x) ! 0 and ln(cos 3x) ! 0

4


ln a ¡ ln b 1 ³a´ = ln provided a > 0, b > 0 b =

2 As x !

¼¡ , 2

)

sin 5x cos 3x sin 3x cos 5x ³ sin 5x ´ ³ 3x ´ ³ cos 3x ´ 1 = lim 53 5 5x sin 3x 3 cos 5x x!0+ ³ sin 5x ´ ³ 3x ´ = 25 £ lim £ lim £ 9 5x 5x!0+ 3x!0+ sin 3x

1 , 1


tan x sec2 x = lim ¼ ¡ sec x tan x sec x x!

lim

¼¡ x! 2

lim

sec x tan x

lim

sec x tan x sec2 x

¼¡ x! 2

=

¼¡ x! 2

= b

lim

¼¡ x! 2

ln(sin 2x) is not defined when sin 3x < 0 ln(sin 3x) ) ¼ 2

and

tan x sec x

1 1

£1£1£1

that is, when ¼ 6 3x 6 2¼

fusing l’Hôpital’s Rule againg =

25 9 25 9

=

2

=

5 3

lim

x!0+

tan x ! 1 and sec x ! 1

The limit has type


¡5 sin 5x ln(cos 5x) cos 5x ) lim = lim x!0+ ln(cos 3x) x!0+ ¡3 sin 3x cos 3x which is ³ ´³ ´

cos x

x!0

0 , 0

)

¼ 3

6x6

y

2¼ 3

S

lies in this interval

lim

¼¡ x! 2

ln(sin 2x) ln(sin 3x)

A

¼

2¼ x T

DNE

C

which is back to where we started. So, l’Hôpital’s Rule was no use here. sin x

tan x cos x = = sin x provided cos x 6= 0 1 sec x ¼ cos x that is, x 6= 2 tan x = lim sin x = 1 ¼¡ sec x x!

Note:

)

lim

¼¡

x! 2

5

1 1 sin x ¡ x ¡ = x sin x x sin x ³1 ´ ³ sin x ¡ x ´ 1 ) lim ¡ = lim sin x x sin x x!0+ x x!0+ a

The limit has type

2

) 3 As x ! 0,

¡ arccos x ¡ x !

¼ 2 ¼ 2

that is,

lim

¼ 2

0 , 0

x!0

= lim

x!0

= lim

1¡x2 3x2

x!0+

The limit also has type )

lim

p

3x2 3x2

1 6

1 ¡ x2

Ã

1 ¡ x2

p

p

3

=

x!0+

1+ 1+

p

1 ¡ x2

b

p

1 ¡ x2

1 ¡ (1 ¡ x2 )

1 ¡ x2 (1 + 1

1 ¡ x2 (1 +

¡

x!0+

0 , 0

´ 1

sin x

p

1 ¡ x2 )

¡ sin x cos x + [cos x + x(¡ sin x)] ¡ sin x = lim + 2 cos x ¡ x sin x x!0 0 = 2¡0 = lim

x!0+

1 1 sin 2x ¡ x ¡ = x sin 2x x sin 2x ³1 ´ 1 sin 2x ¡ x ) lim ¡ = lim sin 2x x!0+ x x!0+ x sin 2x The limit has type

p

)

1 ¡ x2 )

yellow

lim

x!0+

³1

x

¡

0 , 0


1 sin 2x

´

= lim

x!0+

95

2 cos 2x ¡ 1 sin 2x + x(2 cos 2x)

2(1) ¡ 1 0+0

the limit DNE.

100

50

75

25

0

5

95

100

50

75

25

0

5

95

50

75

25

100

magenta

cos x ¡ 1 sin x + x cos x

so we again use l’Hôpital’s Rule.

=

5

0

x

= lim

=0

!

)

cyan

³1

´

¡1

p

1¡

1 3(1)(2)

95

lim

3


0+ p 1

=

100

50

75

25

0

5

x!0


1 ¡ x sin x

x3

x!0

= lim

¡0

0 , 0

¡ arccos x ¡ x

x!0

= lim

¼ 2

¡ arccos x ¡ x ! 0 and x ! 0


¡

¼ 2

³1

black



156

WORKED SOLUTIONS c sec2 x ¡ tan x =

1 sin x ¡ cos2 x cos x

=

1 ¡ sin x cos x cos2 x 1¡

=

1 2

x!1

x!1

1 ¡ 12 (1) 0

= lim

x!1

the limit DNE.

=1

1 , 1


lim

x!1

1+

b


xk kxk¡1 = lim x x!1 e ex

And, by using l’Hôpital’s Rule repeatedly = lim

x!1

³

) )

k(k ¡ 1)xk¡2 ex

1+

lim

x!1

1 x

@ 1+x

1 1 + x¡1 1 fx¡1 = ! 0g x

1 ln =e x

1+

1

1 x

c

ex

lim

³

´x

¡

x!1

fk is fixed, e ! 1g x

1+

x!1

³

x

= lim

x !1 a

ffrom ag

20 1xa 3a 6@1 + ³ 1 ´ A 7 4 5 x

=e

9 x = eln x

x!1

ffrom bg

= esin x ln x ln x ¡1

= e [sin x]

1 kxk

As x ! 0+ , ln x ! ¡1 and [sin x]¡1 ! 1 The limit has type

) for large x, xk is very much greater than ln x ) ln x increases more slowly than any fixed power of x.

a As x ! 1,

³

ln 1 +

1 x

´

1 that is, lim ln 1 + x!1 x Consider x ln 1 +

³

´

ln x x lim = lim x!0+ [sin x]¡1 x!0+ ¡[sin x]¡2 cos x

)

= lim

x!0+

=0


´

magenta

yellow

95

50

75

25

0


=0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100 cyan

0 , 0

¡[sin x]2 x cos x

ln x ¡2 sin x cos x lim = lim x!0+ [sin x]¡1 x!0+ cos x + x(¡ sin x) ¡2(0)(1) = 1¡0

1 !0 x The limit has type 0 £ 1, so we can use l’Hôpital’s Rule.

As x ! 1, ln 1 +

so we can use l’Hôpital’s Rule. 1

´ 1

x

1 , 1

! ln 1 = 0

³

³

.... (1)

ln x Now consider as x ! 0+ [sin x]¡1

=0

50

a

) xsin x = (eln x )sin x


1 ln x x lim = lim x!1 xk x!1 kxk¡1

75

¢´

x!1

100

1 , 1

= lim

25

1

lim

The limit has type

0

¡

a

xk

ln x , k 2Z+ x!1 xk As x ! 1, ln x ! 1 and xk ! 1

5

¢

lim x ln 1+ x

a

8

1

x ln 1+ x

= lim e

20 1xa 3a 1 6 7 = lim 4@1 + ³ ´ A 5 x!1 x

´ a x

b The result in a implies that for large x, is greater than for any fixed k 2 Z + . ) ex increases more rapidly than any fixed positive power of x.

)

¢

=e =e

ex

7 Consider

fa = eln a g

1

k!x0

=0

¡

¢

x ln 1+ x

=e

.. . x!1

1

1+ x

=e

k(k ¡ 1)(k ¡ 2)xk¡3 = lim x!1 ex

= lim

A

¡x¡2

´x

³

´

1

¡x¡2

¡

a For all k 2 Z + , as x ! 1, xk ! 1 and ex ! 1.

6

0

¡1

= lim

¼¡

1 x

ln(1 + x¡1 ) x¡1

= lim

cos2 x

x! 2

)

x!1

sin 2x

lim (sec2 x ¡ tan x) =

)

³

lim x ln 1 +

Now

black



WORKED SOLUTIONS ln x ¡1

x!0+

x!0+

10 x

1

=

(eln x ) x

=e

2 x

= e0 =1 1 x

y

-2_Qw

lim x sin x = lim e [sin x]

Hence,

157

f(x) = -2x - 5, -2_Qw 6 x 6 -1

ln x x

f(x) = x2 - 4, -1 6 x 6 2

(-1, -3)

ln x . x As x ! 1, ln x ! 1 lim

Consider

x!1

The limit has type

1 , 1

½


f 0 (x)

1

)

lim

x!1

)

ln x x = lim =0 x!1 1 x

lim e

ln x x

f (¡1 12 ) = f (2) = 0 and f (x) is continuous and differentiable on [¡2 12 , 2].

lim x x = 1

2

a

f (x) = 3x3 + 5x2 ¡ 43x + 35 on [¡5, 2 13 ]

a

Rolle’s Theorem applies and f 0 (c) = 0 , c = 0 which lies in ]¡2 12 , 2[ .

)

x!1

EXERCISE F 1

f (¡5) = f (2 13 ) = 0

f 0 (x) = 9x2 + 10x ¡ 43 for all x 2 R ) f (x) is continuous and differentiable for all x 2 [¡5, 2 13 ] and f (¡5) = f (2 13 ) = 0 ) Rolle’s Theorem applies p ¡5 § 2 103 f 0 (c) = 0 , c = 9

i f (x) = (x ¡ 1)(x ¡ 2)(x ¡ 4)(x ¡ 5) has zeros: 1, 2, 4, and 5 As f (x) is a real polynomial, continuous and differentiable for all x 2 R , by Rolle’s Theorem, there exist zeros of f 0 (x) in the intervals ]1, 2[ and ]2, 4[ and ]4, 5[ . So at least 3 real zeros exist for f 0 (x). ii f (x) has sign diagram and graph:

+

both of which lie in ]¡5, 2 13 [ b f (x) = jxj ¡ 5 on [¡5, 5]. y

-5

5

If x > 0, If x < 0,

x

0 f+ (x) = 1

b

0 f¡ (x) = ¡1

Thus f (x) is not differentiable for all x 2 [¡5, 5] Rolle’s Theorem does not apply. 1 on [¡ 12 , 7]. x+1

-

is continuous on [¡ 12 , 7] 1 f 0 (x) = exists for all x 2 R , x 6= ¡1 (x + 1)2

Rolle’s Theorem does not apply.

magenta

yellow

95

50

75

25

0

5

95

100

50

75

25

0

on [¡2 12 , 2]

5

95

x < ¡1 x > ¡1

100

50

75

25

0

5

95

100

50

75

25

0

5

cyan

¡2x ¡ 5, x2 ¡ 4,

+ 1

x

2

+ 3

x

i f (x) = (x ¡ 1)2 (x2 + 9)(x ¡ 2) has two real zeros: 1 and 2. As f (x) is a polynomial, continuous and differentiable for all x 2 R , by Rolle’s Theorem, there exists a zero of f (x) in ]1, 2[ . So at least one zero exists for f 0 (x).

100

c

f (¡ 12 ) = 0 and f (7) = 1 78 6= 0

½

+ 5

As y = f (x) has 4 turning points, f 0 (x) has 4 distinct real zeros, 3 guaranteed by Rolle’s Theorem and x = 1.

is differentiable on [¡ 12 , 7]

d f (x) =

+ -3

)

)

4

i f (x) = (x ¡ 1)2 (x2 ¡ 9)(x ¡ 2) = (x + 3)(x ¡ 1)2 (x ¡ 2)(x ¡ 3) which has zeros: ¡3, 1, 2, and 3. As f (x) is a polynomial, continuous and differentiable for all x 2 R , by Rolle’s Theorem, there exist zeros of f 0 (x) in the intervals ]¡3, 1[ and ]1, 2[ and ]2, 3[ . So at least 3 zeros exist for f 0 (x). ii f (x) has sign diagram and graph:

f (x) is continuous for all x 2 R , x 6= ¡1

)

+ 2

We see that there are 3 turning points ) there are exactly 3 real distinct zeros of f 0 (x).

-5

c f (x) = 2 ¡

1

f (x) is not differentiable at x = 0.

½

)

x < ¡1 x > ¡1

) f 0 (¡1) = ¡2

1

)

¡2, 2x,

0 0 ) f¡ (¡1) = f+ (¡1) = ¡2

= e0 = 1

x!1

=

black

Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\157IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 10:26:40 AM BRIAN


158

WORKED SOLUTIONS f (x) is continuous for all x > 0 and f 0 (x) exists for all x>0 Thus, on [49, 51] f (x) is continuous and differentiable. Hence, by the MVT, there exists c 2 [49, 51] such that p p 51 ¡ 49 = f 0 (c)(51 ¡ 49) p 1 ) 51 ¡ 7 = 2 £ p

ii f (x) has sign diagram and graph:

-

-

+

1

f 0 (x) has exactly 2 real distinct zeros, the one guaranteed by Rolle’s Theorem and the one at x = 1.

)

3

2

)

a f (x) = x3 is continuous on [¡2, 2] and differentiable on ]¡2, 2[ .

p

2 c

1 51 ¡ 7 = p c

b

y

Thus, by the MVT, there exists c in [¡2, 2] such that f (2) ¡ f (¡2) = f 0 (c)(2 ¡ ¡2) ) 8 ¡ (¡8) = f 0 (c) £ 4 ) f 0 (c) = 4

1_ _ _ _ f(x) = %% ~`x_ _

, c = § p2 3 p f (x) = x ¡ 2 is continuous on [3, 6] and differentiable on ]3, 6[ .

y

(6, 2)

x

by the MVT, there exists c on [3, 6] such that f (6) ¡ f (3) = f 0 (c)(6 ¡ 3)

1 8


2, p = 1:5 < p 7 x x7 + 1 x

x

9 g(x) =

f (t) dt, x 2 [0, 4]. 0

Z

1

a g(1) =

f (t) dt

b

Z

0

=

³1 + 3´ 2

=

farea of trapeziumg

d dx

Z

f 0 (x),

g 00 (3)

f 0 (3)

= ) = ftangent at (3, 1) is horizontalg

d ) F (x) = dx

µZ

d = dx d du

¶

2

f (t ¡ 2) dt 1

µZ

¶

u

f (t ¡ 2) dt 1

µZ

¶

u

f (t ¡ 2) dt 1

du dx

Z 1

= f (4 ¡ 2) £ 5

Z

= ¡5

3

1 dx converges x4

1

1

converges for all x > 1

Z

magenta

1

1 1

Z

1

1

1 dx x4

¾

1 dx converges for p > 1 xp

+1 dx converges x4 + 1

)

1 dx + x2

1 dx converge x4

x2

fComparison Testg

2

) e¡x 6 e¡x 2

) 0 6 e¡x 6 e¡x

Z

) 06

50

75

25

0

1

1

Z

1 dx and x2

1 1

Z

b For x > 1, x2 > x ) ¡x2 6 ¡x

5

95

100

yellow

x2 + 1 dx < x4 + 1

½Z

1

2

¡x

e 1

50

95

50

75

100

1

25

0

5

95

100

1

1

x dx 2x5 + 3x2 + 1

75

by the Comparison Test

cyan

1

where

¾

25

Z

converges.

50

Z

1 dx converges for all p > 1 xp

0

1

1

75

1

1

5

½Z


x2 + 1 x2 + 1 1 1 < = 2 + 4 4 x +1 x4 x x

Z

x x 1 < < 4 for all x > 1 5 2 5 2x + 3x + 1 2x x x 1 < 4 for all x > 1 ) 0< 2x5 + 3x2 + 1 x Z 1 Z 1 x 1 dx < dx ) 0< 2x5 + 3x2 + 1 x4 1 1 1

¾

1

sin x dx converges x3

) 0
1.

Z

dx

1 dx converges x3

½Z

)

b Assumptions:

where

¾

x2 ¡ 1 dx is a positive constant p x7 + 1

fComparison Testg

= 5 f (2)

5

1

)

) F 0 (3) = f (g(3) ¡ 2) £ g 0 (3)

)

1

1

fChain ruleg

fComparison Testg

1

where

= f (g(x) ¡ 2) g 0 (x)

a

2

1

Z

where u = g(x)

¾


¯ ¯ ¯ sin x ¯ 1 For all x > 1, 0 6 ¯ 3 ¯ 6 3 x x Z 1¯ Z 1 ¯ 1 ¯ sin x ¯ dx 6 ) 06 ¯ x3 ¯ x3

= f (u ¡ 2) g 0 (x)

1

½Z

1

g(x)

0

dx

x2 ¡ 1 dx converges p x7 + 1

2

f (t ¡ 2) dt 1

=

1

g(x)

F (x) =

a

1

x2 ¡ 1 dx converges p x7 + 1

)

=0

1 x1:5

1

1

1

1

g 0 (1) = f (1) = 3

g 00 (x)

½Z

)

fFTOCg

Z

10

Z

=1

0

95

)

= f (3)

f (t) dt

= f (x)

1

g (3)

x

1

dx
2g

Z

x2 ¡ 1 x7 + 1

where

¼ 4

1 x1:5

p

1

Z

£ 2 + 14 ¼(1)2

1

Z

1

dx 6

e¡x dx

¡1

100

=

e

=6+

g (x)

c

2

1

) 0
0 for all x2R,

1 dx x

du dx dx

=

1

1 1

1

=

¡e¡x dx

2 e

=

0
0

0

2

·

1

0

c Pn is:

Z

1

x3 e¡x dx = 6 = 3!

Z

= ¡x2 e¡x + 2

Z

x2 e¡x dx = 2 = 2!

(1) P0 was proved in a.

2 ¡x

x2 e¡x dx = ¡x2 e¡x ¡

)

´

x e

v=x v 0 = 2x

Z

1

0

2

If u = e , u = ¡e¡x ,

1

x1 e¡x dx = 1 = 1!

(2) Assuming the truth of Pk ,

1

If n = 2, we need to find 0

+1

¡eb

b!1

1

¤b 0

¶

x0 e¡x dx = 1 = 0!

So, we predict

¡e¡x (x + 1)

0

¡(b3 + 3b2 + 6b + 6) ¡ (¡6) eb

0

b

xe¡x dx = lim

b!1

1

0

v=x v0 = 1

¸b

=6

b

Z

)

µ

= lim

xe¡x dx

xe¡x dx = ¡xe¡x ¡

= ¡e

¡(x3 + 3x2 + 6x + 6) ex

b!1

Using Integration by Parts with u0 = e¡x , u = ¡e¡x ,

¡x

·

b!1

¡ (¡1)

0

Z

x3 e¡x dx

= lim

=1¡0 =1

Z

1

)

e¡x dx

b!1

0

Z

¶

b

e¡x dx = lim

black

Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\165IB_HL_OPT-Calculus_an.cdr Friday, 8 March 2013 12:20:41 PM BRIAN

¡2x ex2


166

WORKED SOLUTIONS 2

For x > 0, ¡2x < 0 and ex > 0 ) f 0 (x) < 0 )

f (x) is decreasing for x > 0. 1 P

b U=

f (i) =

i=0 1

P

L= 1 P

f (i + 1) =

Z

2

1 P

e an = ¡(i+1)2

=

e

1

2

e¡x dx
0 1 P

1 P i=1

¡1 < i2

1

)

EXERCISE J.1 As n ! 1, n + n3 ! 1 ) )

lim an = 0

´

) an =

lim an = 0 3¡ ¡ 5n = 2 5n2 + 2n ¡ 6 5+ n ¡

lim an =

n!1

=

6 n2

)

)

cyan

magenta

yellow

50

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

+

n2 + 1 + n + 1 ¡ n2

fdividing each term by

p ng

fan g converges to 1. cos2 n 2n

¯ ¯ ¯ cos2 n ¯ ¯ ¯ ¯ 2n ¯ 6

1 , n 2Z+ 2n

)

1 !0 2n jan j is convergent to 0 fSqueeze Theoremg

)

fan g converges to 0.

As n ! 1,

2 n2

+ n12 + n13

5

95

1+

1 n

n2 + 1 + n

n2

p

Now 0 6

+ n2

1+

p

n2 + 1 ¡ n

fan g is divergent.

d an =

+ 2n n3 + n2 + n + 1

100

50

75

25

0

5

=

!

n

(n2 + 2n)(n2 + 1) ¡ n3 (n + 1) (n + 1)(n2 + 1)

1 n

! Ãp

n2 + 1 + n

1

1 As n ! 1, p ! 0 n

n4 + 2n3 + n2 + 2n ¡ n4 ¡ n3 = n3 + n2 + n + 1 =

+1¡n

1

n(n + 2) n3 ¡ 2 n+1 n +1

n3

16 81

=

1¡ p p n¡1 n = c an = p 1 n+1 1+ p

3 5

75

d an =

3

n2 + 1 + n p and as n ! 1, n2 + 1 ! 1 =

5 2 6 ! 0, ! 0, and 2 ! 0 n n n

As n ! 1, )

p

n!1

c an =

n2

p

=

5 n

¡ 2 ¢4

1

p Ã

3n2

¶4

1 n! = (n + 3)! (n + 3)(n + 2)(n + 1)

an =

b

n+1 n ³ ´ 1 = ln 1 + n As n ! 1, an ! ln 1 )

3+

3 n 7 n

) fan g converges to 0.

b an = ln(1 + n) ¡ ln(n)

³

2¡

7 ¡3 ! 0, and !0 n n

lim an =

n!1

= ln

µ =

As n ! 1, (n + 3)(n + 2)(n + 1) ! 1 ) an ! 0

1 !0 n + n3

95

1

3n + 7

n!1

a an =

2

1 a an = n + n3

³ 2n ¡ 3 ´4

As n ! 1,

1 P ¡1 ¡1 dx < x2 (i + 1)2 i=1

1

lim an = 0

n!1

100

c

Z

p p n ! 1, n+1!1

f an =

1 P ¡1 ¡1 , L= 2 2 (i + 1) i=1 i

i=1

p ¢ n+1¡ n

i=0

) f 0 (x) > 0 for all x > 0

b U=

p ¶ n+1+ n p p n+1+ n

µp

¡p

n+1¡n p n+1+ n

1

0

p p n+1¡ n

= p

2

e¡i

f (x) = ¡ 2 = ¡x¡2 x

a

n!1

i=0

e¡(i+1)
0, 2 +

sin_µ µ

)

µ 1 x

1 n

)

jan j is convergent to 0 fSqueeze Theoremg

v u u 3 t

=

Thus j(an + bn ) ¡ (a + b)j < " for all n > N )

¡ n14 + n46 6

1 + n32 + n34 + n16 1

=0

fan g is convergent to 0.

)

1 2 3 n + 2 + 2 + :::: + 2 n2 n n n 1 + 2 + 3 + :::: + n = n2

3 an =

½

n(n+1) 2 = n2

= =

1 2 1 2

³

n

1 1+ n

As n ! 1,

n P

¾

) )

an !

1 2

)

n+1>n

magenta

provided

we require n ¡ 1 >

¯ ¯ 3n + 5 ¯ 7n ¡ 4 lim

n!1

¡

3 7

¯ ¯ N = 1 +

lim an = a and

1 "

lim bn = b,

n!1

lim ¯bn = ¯b

n!1

lim (®an + ¯bn )

Hence,

n!1

³ 1 ń

= ® lim an + ¯ lim bn

fas ®, ¯ are constantsg

n!1

n!1

= ®a + ¯b lim (an ¡ bn ) = a ¡ b

n!1

yellow

50

25

0

=0

5

95

1 n+1

ń

For ® = 1, ¯ = ¡1,

100

lim

fLimit Lawg

n!1

1 nn ³

n!1

= lim ®an + lim ¯bn n!1

n

50

)

95

50

75

1 !0 nn

25

0

5

95

100

50

75

25

0

n+1

0,

n!1

j(an + bn ) ¡ (a + b)j

¡ n2

2 n

does not exist.

= j(an ¡ a) + (bn ¡ b)j

95

3

=

ń

lim an = a and

But

2n5 ¡ n2 + 4 n2 + 1

75

r

1 n

for all n 2 Z +

all n 2 Z + .

p 3

3

n!1

2+

> 2n

there exists N such that jan ¡ aj
2 n

100

But

¯ ³ ´¯ ¯sin 1 ¯ 6 ¯ ¯ n

y 1

167

black



168

WORKED SOLUTIONS

EXERCISE J.2 1

a

ii Let cn =

i un =

2n ¡ 7 3n + 2

en

en+1

2(n + 1) ¡ 7 2n ¡ 7 = ¡ 3(n + 1) + 2 3n + 2 2n ¡ 5 2n ¡ 7 = ¡ 3n + 5 3n + 2 =

(2n ¡ 5)(3n + 2) ¡ (2n ¡ 7)(3n + 5) (3n + 5)(3n + 2)

=

6n2 ¡ 11n ¡ 10 ¡ [6n2 ¡ 11n ¡ 35] (3n + 5)(3n + 2)

=

25 (3n + 5)(3n + 2)

=

(1 ¡ e)(e3n+1 + en ) (e2n+2 ¡ 1)(e2n ¡ 1)

) fun g is monotonic (increasing)

Also, cn > 0 for all n 2 Z +

i If bn =

Now as n ! 1, en ! 1, and e¡n ! 0 1 ) !0 en ¡ e¡n 1 ) lim n =0 n!1 e ¡ e¡n

=

3n+1 + 32n+1 ¡ 3n ¡ 32n+1 (1 + 3n+1 )(1 + 3n )

=

3n (3 ¡ 1) (1 + 3n+1 )(1 + 3n )

1 £ 3 £ 5 £ 7 £ :::: £ (2n ¡ 1) 2n n! un+1 1 £ 3 £ 5 £ 7 £ :::: £ (2n ¡ 1)(2n + 1)2n n! = un 1 £ 3 £ 5 £ 7 £ :::: £ (2n ¡ 1)2n+1 (n + 1)!

Let un =

2 )

> 0 for all n 2 Z + +

for all n 2 Z + .

) fun g is bounded

1 !0 1 + 3n

Proof: (by induction)

p

(1) x2 = 4 + 3(0) = 2 ) x2 > x1 ) P1 is true.

cyan

magenta

yellow

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

n!1

5

95

lim bn = 1

100

)

a Pn : fxn g is monotonic increasing, that is, xn+1 > xn for all n 2 Z +

95

From (1), as n ! 1,

is its greatest lower bound.

Thus fun g is convergent. p 3 x1 = 0 and xn+1 = 4 + 3xn , n 2 Z +

6 bn < 1 for all n 2 Z +

) fbn g is bounded

1 2

Also un > 0 ) 0 < un < 1

) bn < 1 for all n 2 Z + 3 4

2n + 1 2n + 2

Thus u1 =

3n 1 + 3n ¡ 1 Also bn = = 1 + 3n 1 + 3n 1 ) bn = 1 ¡ .... (1) 1 + 3n So,

=

100

3 4

2n + 1 2(n + 1)

) un+1 < un for all n 2 Z + fun > 0g ) fun g is monotonic (decreasing)

3 4

Hence its smallest member is b1 = bn >

=

< 1 for all n 2 Z +

) fbn g is monotonic (increasing) )

e for all n 2 Z + e2 ¡ 1

As fcn g is monotonic and bounded ) fcn g is convergent.

3n+1 3n ¡ n+1 1+3 1 + 3n

) bn+1 > bn for all n 2 Z

fen > 0, e2n > 1g

) fcn g is bounded

2n ¡ 7 3n + 2 < 3n + 2 3n + 2

3n , 1 + 3n

bn+1 ¡ bn =

0 < cn 6

Thus

So ¡1 6 un < 1 ) fun g is bounded.

50

e3n+1 (1 ¡ e) + en (1 ¡ e) (e2n+2 ¡ 1)(e2n ¡ 1)

) fcn g is monotonic (decreasing) e is its least upper bound. Thus c1 = 2 e ¡1

Thus fun g has an upper bound of 1

75

=

cn+1 < cn for all n 2 Z +

)

) un < 1 for all n 2 Z +

25

e3n+1 ¡ en+1 ¡ e3n+2 + en (e2n+2 ¡ 1)(e2n ¡ 1)

> 0 for all n 2 Z + ) un+1 > un for all n 2 Z +

Also, as un =

0

=

Now 1 ¡ e < 0, e3n+1 + en > 0, e2n+2 > 1, and e2n > 1 ) cn+1 ¡ cn < 0 for all n 2 Z +

ii As fun g is monotonic increasing u1 = ¡1 is its greatest lower bound.

5

en

) cn+1 ¡ cn = 2n+2 ¡ 2n e ¡1 e ¡1

un+1 ¡ un

b

1 en = 2n ¡n ¡e e ¡1

black

Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_an\168IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 4:07:29 PM BRIAN


169

WORKED SOLUTIONS (2) If Pk is true, then xk+1 > xk , k 2 Z + .... (¤) 2 2 Now xk+2 ¡ xk+1 = 4 + 3xk+1 ¡ (4 + 3xk ) = 3(xk+1 ¡ xk ) >0 ffrom ¤ g 2 2 > xk+1 ) xk+2 ) xk+2 > xk+1

d Given fun g converges, then lim un = lim un+1 = L, say n!1

n!1

) 2

L ¡L¡1=0

)

) Pn is true fPOMIg

5 un+1 = a

Proof: (by induction) P1 is true (2) If Pk is true then xk < 4 for k 2 Z + 2 Now xk+1 = 4 + 3xk )

fusing Pk g

< 16 fas xk+1 > 0g

)

d From a and c, fxn g is convergent.

r lim xn+1 =

Since

n!1

) )

) 4

lim xn+1 = L also.

n!1

4+3

³

p L = 4 + 3L L2 = 4 + 3L

´

u4 =

3 2 5 3 8 5

)

)

L=

for all n 2 Z + .

1 un

)

´

2 L

fas L > 0g

yellow

1 2

L+

) )

lim un =

n!1

If

p

fas L > 0g 3.

lim un = L then

n!1

)

95

´

3 . un

L

1 2

³

un +

k un

´

f Suppose u1 = a, a > 0, and un+1 =

50

un +

lim un+1 = L also.

then

n!1

75

³

ń!1 3

e If u1 = 6 and un+1 = p lim un = k.

25

0

1 2

³

L2 = 3 p L= 3

)

5

95

100

50

75

25

0

5

95

100

50

75

25

0

2 L

3 2L = L + L 3 L= L

) u1 = 2 and un+1 = 1 +

5

95

100

50

75

L+

lim un+1 = L also.

n!1

2 L

n!1

) fun g is bounded.

25

³

lim un = L then

If

b un is defined by:

magenta

=1 iii u1 = 7 u2 ¼ 3:642 90 = 1:5 u3 ¼ 2:095 94 ¼ 1:416 67 u4 ¼ 1:525 08 ¼ 1:414 22 u5 ¼ 1:418 24 ¼ 1:414 21 ¼ 1:414 21 u6 ¼ 1:414 21 u7 ¼ 1:414 21 ¼ 1:414 21 ¼ 1:414 21 u8 ¼ 1:414 21

d Suppose u1 = a, a > 0, and un+1 =

c fun g is not monotonic as u2 < u1 but u3 > u2 . All terms of the sequence fun g are clearly positive. 1 So, as un¡1 > 0 and un = 1 + , we deduce un¡1 un > 1 for all n 2 Z + . 1 < 1, for n ¡ 1 > 1 Thus un¡1 > 1 and un¡1 ) un 6 2 ) 1 < un 6 2 for all n 2 Z +

cyan

1 2

L2 = 2 p L= 2

)

n!1

u3 =

0

L=

n!1

lim xn = 4 2 1

ii u1 u2 u3 u4 u5 u6 u7 u8

lim un = L then

2L = L +

)

lim xn ,

L ¡ 3L ¡ 4 = 0 (L ¡ 4)(L + 1) = 0 ) L = 4 or ¡1 ) L = 4 fas xn > 0g

u2 =

5

L=

)

2

a u1 = 2 =

=5 = 2:7 ¼ 1:720 37 ¼ 1:441 46 ¼ 1:414 47 ¼ 1:414 21 ¼ 1:414 21 ¼ 1:414 21

L=

1 2

³

k L+ 2 L

1 2

³

un +

lim un+1 = L also.

´

k . un2

´ n!1

100

)

´

n!1

Hence, 0 6 xn < 4, so fxn g is bounded. lim xn = L say, then

2 un

c Yes, and if

) Pn is true fPOMIg

n!1

un +

p 1§ 5 2

b No, as in a where u1 = 1, u2 > u1 but u3 < u2 .

Thus P1 is true and Pk+1 is true whenever Pk is true

If

L=

³

1 2

i u1 u2 u3 u4 u5 u6 u7 u8

(1) If n = 1, 0 < 4 )

) xk+1 < 4

)

p 1+ 5 But L > 0, so lim un = n!1 2

=0 x5 ¼ 3:8752 x9 ¼ 3:9975 =2 x6 ¼ 3:9529 x10 ¼ 3:9991 ¼ 3:1623 x7 ¼ 3:9823 x11 ¼ 3:9996 ¼ 3:6724 x8 ¼ 3:9934 The experimentation suggests that 4 may be an upper bound. p c Pn : If x1 = 0 and xn+1 = 4 + 3xn for all n 2 Z + then xn < 4.

b x1 x2 x3 x4

2 xk+1

1 L

L2 = L + 1

)

fas xk+2 , xk+1 > 0g


2 ) xk+1 < 4 + 3(4)

L=1+

black



170

WORKED SOLUTIONS

)

) )

lim un =

p 3

1+

=1+

1 n

) g(x) > 0 Since f 0 (x) = f (x) g(x) where f (x) > 0, g(x) > 0 for all x > 0, we deduce f 0 (x) > 0 for all x > 0 )

³ ´1 n

+:::: +

³

n

+

³ ´ 1 n ´

n n¡1

+ n2

2

1

+

nn¡1

³1´

en = 1 + n

+

3

n3

n

nn

n

³

µ

::::

´

n ¡ [n ¡ 1] n

³

· 1 ln(1 + x¡1 ) + x

¶ + (x + 1)¡2

d

2

¡ 1 ¢n¡1

)

2

yellow

³

´

1 n =e n!1 n ³ n + 1 ń lim =e n!1 n lim

i )

50

25

0

¡1 1 + x(x + 1) (x + 1)2

95

=

100

¡1 1 + x2 + x (x + 1)2

75

¡ 1 ¢n

fsum of a GSg

Thus 2 6 en < 3 ) fen g is bounded. So, en is monotonic (increasing) and bounded ) fen g is convergent.

=

magenta

1 ¡ 12

) en < 3 for all n 2 Z + , n > 3

x+1 x2 x2

¡ ¢n 1(1 ¡ 12 )

) en < 1 + 2(1 ¡ ) en < 3 ¡

50

µ

1 + x¡1

i 1

25

0

5

95

100

50

75

cyan

¡x¡2 1 + x¡1

iii Thus en < 1 +

95

= f (x) g(x), say.

¸

50

= f (x) ln(1 + x¡1 ) ¡

Now g 0 (x) =

1 1 1 1 + + + :::: + , 2! 3! 4! n! 1 1 1 en < 1 + 1 + 12 + 2 + 3 + :::: + n¡1 2 2 2

So, as en = 1 + 1 +

75

h

25

= 2k Thus P3 is true and Pk+1 is true whenever Pk is true. ) Pn is true fPOMIg

25

= f (x) ln(1 + x

x¡1 )¡ 1 + x¡1

¸

¡x¡2

0

¡1

0

> 2 £ 2k¡1

5

)

·

5

) (k + 1)! = (k + 1)k!

)

95

¡1

= ex ln(1+x

, n>0

Proof: (by induction)

´

100

f 0 (x)

)

¡1

ń

(1) If n = 3, 3! > 22 ) 6 > 4 ) P3 is true (2) If Pk is true, then k! > 2k¡1 , k 2 Z +

, x > 0, x 2 R .

Now f (x) = ex ln(1+x

³

Pn : n! > 2n¡1 for all n > 3, n 2 Z +

¶

´³

75

´x

5

f (x) =

1 1+ x

´

n+1 1 1 > 1+ n+1 n > en for all n > 1

ii Using b, since 2 3 n¡1 1 < 1, 1¡ < 1, 1¡ < 1, 1¡ < 1, ...., 1¡ n n n n 1 1 1 1 en < 1 + 1 + + + + :::: + 2! 3! 4! n! 1 1 < n¡1 for all n > 3, We now need to show that n! 2 n 2 Z + or n! > 2n¡1 .

i From b, en = 1 + 1+ many other positive terms ) en > 2 .... (1) To prove that en+1 > en for all n 2 Z + , consider

³

1+

Hence, from (1), 2 6 en < en+1

1 1 1 1 2 1¡ + 1¡ 1¡ 2! n 3! n n ³ ´ ³ ´ 1 1 n¡1 +:::: + 1¡ :::: 1 ¡ n! n n

) en = 1 + 1 +

³

) en+1

³ ´ 1 n

³ ´ n(n ¡ 1) 1

³n ¡ 1´

1 +:::: + n!

f (x) is strictly increasing for all x > 0

)

³ ´ 1 n

n 2! n2 ³ ´ n(n ¡ 1)(n ¡ 2) 1 + + :::: 3! n3 ³ 1 ´ n(n ¡ 1)(n ¡ 2)::::(n ¡ [n ¡ 1]) + n! nn ³n ¡ 1´ 1 ³n ¡ 1´³n ¡ 2´ 1 ) en = 1 + 1 + + 2! n 3! n 2

c

¡1 x(x + 1)2

x!1

k.

ń

1

¡(x + 1) + x x(x + 1)2

) g 0 (x) < 0 for all x > 0 ) g(x) > lim g(x) = ln 1 ¡ 0

n!1

³

b

=

L3 = k p 3 L= k

a

g 0 (x) =

)

k L2

L=

)

6

k L2

2L = L +

1+

100

)

black



WORKED SOLUTIONS

³

´

1 n n!1 n ³ n ¡ 1 ń = lim n!1 n lim

1¡

= lim

d

n=1 1

P

³ m ´m+1

n=1

freplacing n by m + 1g

m+1

m!1

)

³

lim

m!1

m+1 m

n

n! = nn

)

³n ¡ 1´³n ¡ 2´

n! 6 nn

) )

0

0,

´2 ³ 1 ´

n! lim =0 n!1 nn

)

´

1 converges fp-series testg n2

³

)

n

³

ń ³

P

an2 4n4 + 12n3 + 9n2 n3 = £ 2 7 5+n 4 bn

³3´³2´³1´

n ¡ 2 of these

n! 6 nn

1 P 1 1 ¡ where n n=1 n2

n=1 1

1 ¡ 2 n n

=

³ n ¡ 1 ń¡2 ³ 1 ´ n

1 P

=

1 ³ P 1

)

|

´

1 diverges and n

n! n(n ¡ 1)(n ¡ 2) £ :::: £ 3 £ 2 £ 1 = nn n £ n £ n £ :::: £ n £ n £ n

ii

1 n2

2n2 + 3n 2 2 Let an = p and bn = p 5 + n7 n3

´m

1 = e

1

¡

n=1

m lim m!1 m+1

=

1 ³ P 1

171

black



172 4

WORKED SOLUTIONS a

1

p

n(n + 1)(n + 2)

1 P

1

n=1

n1:5

1 P

)

1 1 = 1:5 and 0

1

p

converges

n(n + 1)(n + 2)

n=1

fComparison testg

1 1 > for all x > 0 ln x x

)

1 1 > for all n 2 Z + , n > 2 ln n n

But

n(n + 1)(n ¡ 1) = n3 ¡ n < n3

b

)

converges fp-series testg

p 3

)

n(n + 1)(n ¡ 1) < n

1 > p 3 n n(n + 1)(n ¡ 1)

)

1 P

)

n=2

where

n=2

1 P

5

c As 0 6 sin2 µ 6 1 for all µ 2 R , 1 sin2 n p 6 1:5 n n n

P

)

n=1

sin2 n p n n

fComparison testg

also converges

) 2

1 ¡ ¢ P 2 n 3

n=1

Hence f

06x
1 1 + c > 1 or 1 + c < ¡1 c > 0 or c < ¡2 p 3¡1 Thus c = 2 x is continuous for all x > 1 .... (1) 7 a f (x) = 2 x +1

converges fComparison testg

y y=x y = ln x 1

1 0 ) x2 + 1 cannot be 0g

x

f 0 (x) =

1(x2 + 1) ¡ x(2x) 1 ¡ x2 = 2 (x2 + 1)2 (x + 1)2

cyan

magenta

yellow

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

But x > 1 ) x2 > 1 ) 1 ¡ x2 6 0

black



173

WORKED SOLUTIONS Thus f 0 (x) 6 0 for all x > 1 ) f (x) is decreasing for all x > 1 .... (2)

) f (x) is decreasing when ln x > 1 ff 0 (x) < 0g that is, when x>e

And as x > 1, f (x) > 0 for all x > 1 ) f (x) is positive for all x > 1 .... (3)

)

ln x is continuous for x > 1 .... (3) x From (1), (2), and (3), the Integral Test applies for x > 3. Also f (x) =

From (1), (2), and (3) we can apply the Integral Test.

Z

1

x dx x2 + 1

Now 1

µZ

b

b!1

1

µ Z 1 2

= lim

b!1

= lim

¶

1

Now 3

ln x dx = lim b!1 x

1 (ln(b2 b!1 2

b!1

which DNE, fas b ! 1, ln(b + 1) ! 1g )

x

2

b f (x) = xe¡x =

2

ex

2 fex

2

Thus

n=3

6= 0 for all x > 1g

2

2

f 0 (x) = 1e¡x + xe¡x (¡2x) = e¡x (1 ¡ 2x2 ) ) f 0 (x) =

1 ¡ 2x2 ex

2

Since

n=1

n=1

2

2x > 1

1 ¡ 2x2

) f 0 (x) = )

1 .... (2)

f 0 (x) =

And f (x) is positive for all x > 1 .... (3)

)

From (1), (2), and (3) we can apply the Integral Test.

Z

1

Now

µZ

2

b

xe¡x = lim

b!1

1

= lim

µZ µ

b

³

)

1 P

2

ne¡n

1

¡ 12

¶

2

³

1

2

e¡b ¡

1 e

1 dx = lim b!1 x ln x

)

´´

µZ

b

2

µZ

b!1

magenta

¶

1 du dx u dx

ln b

£

ln 2

¶

1 du u

¤ ln b

ln u

ln 2

b!1

)

yellow

1 P

95

which DNE fas b ! 1, ln(ln b) ! 1g 1 diverges. n ln n

100

50

75

25

0

n=2

5

95

100

50

25

0

5

¡(ln x + 1) [x ln x]2

= lim (ln(ln b) ¡ ln(ln 2))

1 ¡ ln x = x2

75

x ¡ ln x(1)

=

du 1 fletting u = ln x, = g dx x

= lim

x2

95

1

b!1

100

50

75

25

0

5

95

100

50

75

25

0

5

1 x

= lim

³ ´

cyan

³ ´

[x ln x]2

2

ln x is positive .... (1) x

and f (x) =

1 P ln n ln 1 ln 2 + + , 1 2 n=3 n

0(x ln x) ¡ 1(1 ln x + x

Now

n=1

0

=

1 is positive for x > 2 .... (1) x ln x fln x > 0 for x > 1g

Z

1 2e

1 x

diverges.

f 0 (x) < 0 when ln x + 1 < 0, which is when x > e¡1 0 f (x) is decreasing for x > 2 .... (2)

is convergent.

c For x > 2, f (x) =

[ln b]2 ! 1g 2

From (1), (2), and (3), the Integral Test applies for x > 2.

ib ¶ 2

h

¶

Also f (x) is continuous for all x > 2 .... (3)

¡ 12 e¡x (¡2x) dx

¡ 12 e¡x

b!1

=

)

1

= lim

b!1

2

xe¡x dx

= lim

b!1

¶

n

3

[ln b]2 [ln 3]2 ¡ 2 2

diverges also.

n

d f (x) =

) 1 ¡ 2x2 < 0

n

1 P ln n

1 P ln n

< 0 for all x > 1

Now if x > 1 then

1 P ln n

dx

¸b

[ln x]2 2

which DNE fas b ! 1,

is continuous for all x > 1 .... (1)

x

¶

" " f (x) f 0 (x)

µ

b!1

n diverges fIntegral Testg n2 + 1

n=1

[ln x]

³1´

3

= lim

2

1 P

b 1

= lim

+ 1) ¡ ln 2)

ln x dx x

3

·

1

¶

b

µZ

b!1

¤ b´

= lim

µZ

= lim

¶

2x dx x2 + 1

ln(x2 + 1)

2

b!1

b

1

³ £ 1

Z

x dx x2 + 1

= lim

f (x) is decreasing for x > 3 .... (2)

black



174

WORKED SOLUTIONS 1 is positive for all x .... (1) 1 + x2

)

¡2x (1 + x2 )2

1

1

Z

1

Now 1

1 . 2

.... (¤)

1 dx = lim b!1 1 + x2 = lim

µZ

b

1

£

Z

1

1

1 dx 1 + x2

= So, in ¤,

¼ 4

p = n fas n 6= 0g n p Since 0 6 n 6 Sn , p and lim n DNE,

¤b

1

n!1

by the Comparison test fSn g diverges.

+

) )

P 1

1 P

ln n > 1

)

n ln n > n

) 1 P

an is convergent

n=1

)

)

magenta

yellow

25

0

5

95

100

50

75

25

0

50

75

25

0

5

95

100 cyan

lim

and

3 ¡ 13 (7) 2

2 =

9¡7 2 = = 3£2 3£2

1 3

P1 is true.

12 22 32 42 k2 + + + + :::: + 31 32 33 34 3k 3 ¡ 3¡k (k2 + 3k + 3) = (*) 2

12 22 32 42 k2 (k + 1)2 + 2 + 3 + 4 + :::: + + 31 3 3 3 3k 3(k+1)

)

lim an = 0

1 3

3 ¡ 3¡1 (12 + 3(1) + 3)

(2) If Pk is true, then

1 DNE n!1 an 1 P 1 is divergent ) a n=1 n

75

50

=

=

3 ¡ 3¡k (k2 + 3k + 3) (k + 1)2 + 2 3(k+1)

=

3 ¡ 3¡(k+1) (3k2 + 9k + 9) + 3¡(k+1) 2(k2 + 2k + 1) 2

=

3 ¡ 3¡(k+1) [(3k2 + 9k + 9) ¡ (2k2 + 4k + 2)] 2

n!1

)

25

RHS =

1 P 1 1 1 = p + converges. ln n 2 ln 2 n=3 np ln n

5

1 P

(1) If n = 1, LHS = p

95

n=2

3 ¡ 3¡n (n2 + 3n + 3) , n 2 Z+ 2

Proof: (By the Principle of Mathematical Induction)

1 also converges fComparison testg np ln n np

12 22 32 42 n2 i2 = 1 + 2 + 3 + 4 + :::: + n 3i 3 3 3 3 3 =

1 converges for p > 1 fp-series testg np

P

n P i=1

1 1 < p np ln n n 1 1 P P 1 1 < p p n ln n n n=3 n=3

95

)

n=3 1

the series diverges (by definition).

a Pn : Sn =

12

50

P

)

1 p = lim Sn if this limit exists, n n!1

but from c the sequence of partial sums diverges )

75

n=3 1

p

100

But

n=1

1 diverges fComparison testg np ln n

If p > 1, and n > 3,

)

0

1 diverges. n ln n

np ln n < n ln n

n=2

5

1 P

1 1 > np ln n n ln n 1 1 P P 1 1 > p n ln n n ln n n=2 n=2

)

10

n!1

¼ . 4

n=2

)

1 lim p = 0 (from a), this has no bearing n on the convergence or otherwise of the series.

d Even though

9 If p = 1, we showed in 7 d that If p < 1, np < n

1 if k < ng n

> p

= p

b!1 ¼ ¡ ¼4 2 ¼ 4

1 < 1 + n2

1 n

1

= lim (arctan b ¡ arctan 1) =

1

fsince p

1 dx 1 + x2

arctan x

b!1

1

1 2 3 1 1 1 1 > p + p + p + :::: + p n n n n

.... (3)

1 P 1 1 dx < < a1 + 1 + x2 1 + n2 n=1

where a1 = f (1) =

p i

= p + p + p + :::: + p

From (1), (2), and (3), the Integral Test applies for x > 1. 1

n P 1 i=1

Also f (x) is continuous for all x 2 R

Z

n!1

b Sn =

f 0 (x) < 0 for all x > 1 .... (2)

Now

1 lim p = 0 n

)

) f 0 (x) = ¡(1 + x2 )¡2 (2x) =

p n!1

a As n ! 1,

11

100

8 f (x) =

black


fusing *g


WORKED SOLUTIONS To be correct to 6 decimal places, we require

=

3 ¡ 3¡(k+1) (k2 + 5k + 7) 2

=

3 ¡ 3¡(k+1) (k2 + 2k + 1 + 3k + 3 + 3) 2

=

3 ¡ 3¡(k+1) [(k + 1)2 + 3(k + 1) + 3] 2

1 < 0:000 000 5 3k3 ) 3k3 > 2 £ 106 ) k3 >

b

) k > 88 )

since 2 £ 3n grows much faster than n2 + 3n + 3,

2

1 P

c Yes,

3 n!1 2

n!1 n2

n=1

3n

Hence

n2

n=1

3n

=

n¡1 n¡2 ¡ n+1 n n2 ¡ n ¡ [n2 ¡ n ¡ 2] = n(n + 1) 2 = , n > 2, n 2 Z + n(n + 1)

) an =

3 2

converges, by definition, since from b the 3 . 2

3 . 2

=

a1 + a2 + a3 + a4 + ::::

b

=0+

f (x) dx < Rk

Pn is true fPOMIg

n!1

2 £ 106 3

r

Thus Pk+1 is true whenever Pk is true, and P1 is true. )

175

black



176

WORKED SOLUTIONS m ! 1 ) S2m ! 1 2 Thus Sn is divergent.

1 2 3 k k+1 + + + :::: + + 2! 3! 4! (k + 1)! (k + 2)!

)

(k + 1)! ¡ 1 k+1 + (k + 1)! (k + 2)!

= =

c As m ! 1, 1 +

³ k + 2 ´ µ (k + 1)! ¡ 1 ¶ k+2

(k + 1)!

+

EXERCISE K.3

k+1 (k + 2)!

1

a

(k + 2)! ¡ (k + 2) + k + 1 (k + 2)!

=

1 P n=1

)

n=1

) Pn is true fPOMIg 1 d Since Sn = 1 ¡ , lim Sn = 1 (n + 1)! n!1 1 P n ) =1 (n + 1)! n=1 a

S16 = 1 + +

) S16 > 1 + S21 = 1 + S22 = 1 + S23 = 1 + S24 = 1 +

9

+

1 2

+

¡1

) S16 > 1 + b S20 = 1 +

+

¡1

) S16 > 1 + +

1 2

¢ 1

3

+

1 10

+

1 11

4

+

1 4

¡1

1 P

4

+

5

1 12

+

¢

¡1

+

+

1 7

1 13

+

1 14

+

1 8

+

1 8

+

1 16

+

¡1 8

1 1 1 + 16 + 16 + 16 16 1 + 12 + 12 + 12 2 4 2

0 2 1 2 2 2 3 2 4 2

1 6

+

+

n=1 n

=

+

+ 1 16

1 15

1 8

+

n=1 n 1

+

¢

1 16

1 16

+

P 1 1

)

¢

n=1 1

n=1

=

2

1 P 1

=

Proof: (by induction on m) (1) If m = 0, S20 = S1 = 1 + 02 = 1 X ) P0 is true k (2) If Pk is true, S2k > 1 + 2 Now ³ ´ 1 1 1 S2k+1 = S2k + + :::: + k + k k¡1 2 +1 2 ¡1 2

k + 2

1 1 1 + + :::: + 2k 2k 2k

|

{z

n

) S2k+1 ) S2k+1

¡

diverges.

n2

P n¡1

¡

n2

1 P 1

n=1

n

¡

1 P 1 n=1

¸

n2

which converges fp-series testg

1 P (¡1)n

ln n

where bn =

1 ln n

1 P

(¡1)n bn

n=2

Consider y = ln x, x > 0, x 2 R y

¶

ln x is increasing for all x 2 R , x > 0. y = ln x

}

1

) x

1 is decreasing ln x for all x 2 R , x > 0

1 is decreasing for all n > 2, n 2 Z + ln n lim bn = 0

Thus bn = Also

n!1

)

magenta

yellow

1 P

95

(¡1)n converges ln n

100

50

75

25

0

5

95

100

50

75

25

0

5

95

n=1 1

2k

100

50

75

25

0

5

95

100

50

75

25

n2

1 P 1¡n

·

n2

n=2

0

n

which is an alternating series of the form

) Pn is true fPOMIg

5

n=1

P 1

n=1

1 P 1

n=2


cyan

¡

converges

2k¡1 of these

) S2k+1

n2

¸

| {z }

³1´ k¡1

k >1+ +2 2 k > 1 + + 12 2 k+1 >1+ 2

n=1 1

1 P 1

1 1 1 1 ¡ + ¡ + :::: ln 2 ln 3 ln 4 ln 5

a

m , m 2 Z , m > 0. 2

n

n2

1 P 1

n=1

¡

n

P 1

c

n=1

) S2k+1 > 1 +

¡

diverges

=

µ

¡

P 1

n=1

So, we conjecture: S2m = 1 +

·

n=1

| {z }

¢

1 16

1 P 1¡n

¡

n=1 n

8

+

1 P 1

=2

¢ 1

n=1

1 converges n2

1¡n diverges. n2

1 P 1

b

1 P

1 diverges fp-series testg n

n=1


4

1 P

and

(k + 2)! ¡ 1 = (k + 2)!

¡1

1 1 P P 1¡n 1 1 = ¡ where 2 2 n n n n=1 n=1

black



WORKED SOLUTIONS 1 p 2 x n , consider f (x) = b For (¡1)n¡1 n+4 x+4 n=1

1 P

1 1 x¡ 2

where f 0 (x) = 2

(x + 4) ¡

1 x 2 (1)

p

f 0 (x) 6 0 for all x > 4 f (x) is decreasing for all x > 4 p n then bn is decreasing for all n > 4 Thus if bn = n+4

d

n!

n

³

1 = 1+ n

1 P

Hence,

e

an is divergent fRatio Testg

³¼ ´

n=1

b Consider f (x) = sin f 0 (x) = cos

x

³¼ ´ x

£

n!1

for all x > 1

¡¼ = x2

³¼´

(¡1)n sin

n

n=2

¡¼ cos

³¼´

Hence

x

cyan

f Let f (x) =

which is < 1

(¡1)n is convergent fRatio Testg 2n n!

x2 +1

x3

) f 0 (x) =

and for n > 2, 2 ¡ x3 6 0

yellow

50

75

25

0

5

95

100

50

75

95

magenta

is convergent.

50

n

75

(¡1)n sin

25

³¼´

25

0

5

100

95

1 P n=1

fAlternating Series Testg

n=1

50

¯ ¯ ¯ an+1 ¯ = 0 ¯ ¯ n!1 a lim

n

is convergent

0

)

75

is convergent fRatio Testg

(¡1)n 2n n! ¯ ¯ a 1 2n n! ¯ n+1 ¯ = ) ¯ £ ¯ an 2n+1 (n + 1)! 1 1 = 2(n + 1)

x2

5

P 1

100

1 P

Hence,

25

(2n ¡ 1)!

2x(x3 + 1) ¡ x2 (3x2 ) (x3 + 1)2

=

2x4 + 2x ¡ 3x4 (x3 + 1)2

=

x(2 ¡ x3 ) (x3 + 1)2

lim bn = 0

Also

0

(2n ¡ 1)!

(¡1)n¡1

Let an =

)

) f 0 (x) 6 0 for all x > 2, x 2 R ³¼ ´ ) bn = sin is decreasing for all n > 2, n 2 Z + n

5

1 P (¡1)n¡1

fAlternatively the Alternating Series Test can be usedg

which is > 1

n

1 1 1 1 1 ¡ + ¡ + ¡ :::: 1! 3! 5! 7! 9!

n!

n=1

95

)

n! nn

ń

¯ ¯ ¯ an+1 ¯ = e ¯ ¯ a n!1 lim

=

n!

)

f 0 (x) 6 0 for all x > 2

)

f (x) is decreasing for all x > 2

100

)

´

Let an =

(¡1)n nn

=

which is < 0 for all x > 2

¯ ¯ 1 (2n ¡ 1)! ¯ an+1 ¯ = ) ¯ £ ¯ an (2n + 1)! 1 ¯ ¯ an+1 ¯ 1 ¯ ) ¯ ¯= an (2n + 1)(2n) ¯ ¯ ¯ an+1 ¯ = 0 which is < 1 Thus lim ¯ ¯ n!1 an ³ n¼ ´ 1 sin P 2

p (¡1)n¡1 n is convergent n+4

£

n¼ 2

n=1

fAlternating series testg

)

x

(¡1)n¡1 is convergent. p 3 ln n

=

n!1

¯ ¯ ¯ an+1 ¯ = (n + 1)n+1 ¯ ¯ an (n + 1)! ³ n + 1 ń

³

P sin n=1

lim bn = 0

Let an =

1 P

n=2

1

1 p n !0 Also, as n ! 1, bn = 4 1+ n

a

4 3x(ln x) 3

³1´

fAlternating Series Testg

) )

3

¡1

lim bn = 0,

4¡x 2 x(x + 4)2

n=1

0

n!1

= p

1 P

4

) f (x) is decreasing for all x > 2 1 Hence bn = p is decreasing for all n > 2 and since 3 ln n

x 2 = (x + 4)2

)

= [ln x]

) f (x) = ) f (x) =

p ¡ x

x

)

¡3

ln x

(x + 4)2 p2 ¡

1

1

¡ ¡ 13 [ln x] 3

0

(x + 4)2

1p p2 2 x+ x

=

c Consider f (x) = p 3

177

black



178

WORKED SOLUTIONS 1 P

So, if

(¡1)n+1

n=1

n2 = n3 + 1

1 P

So, by the Estimation Theorem, jS ¡ S4 j 6 b5 ) S4 ¡ b5 6 S 6 S4 + b5 ) 0:841 465 6 S 6 0:841 471

(¡1)n+1 bn then bn

n=1

is decreasing for all n > 2 and also

1 n

lim bn = lim

n!1

n!1

1 P

Hence, )

(¡1)

n+1

(¡1)

n+1

n=1

n=1

n!

As 0
0, fbn g is

yellow

a S2 = (¡1)0 b1 + (¡1)1 b2 = b1 ¡ b2 But bn+1 6 bn for all n 2 Z + ) b2 6 b1 )

S2 > 0

50

b S4 = (¡1)0 b1 + (¡1)1 b2 + (¡1)2 b3 + (¡1)3 b4 = b1 ¡ b2 + b3 ¡ b4

25

0

lim bn = 0.

n!1

5

95

100

50

75

25

0

5

95

50

100

magenta

¡

n=1

is convergent

75

25

0

5

95

cyan

1 384

¼ 0:000 021 70 < 0:000 05

=1 = 0:875 ¼ 0:912 037 0 ¼ 0:896 412 0 ¼ 0:904 412 0

75

n=1

100

50

75

25

0

1 1

6 We are given S =

fAlternating Series Testg 1 1 1 1 1 Now S = ¡ + ¡ + ¡ :::: 1! 3! 5! 7! 9! 1 and b5 ¼ ¼ 0:000 002 76 < 0:000 05 9! and S4 ¼ 0:841 468 25

5

(¡1)n¡1 = n3

lim bn = 0

P

+

An estimate of the error in using S10 to approximate S is b11 = 1113 ¼ 0:000 751 3.

n!1

Thus

1 P

5

n=1

(¡1)n¡1

1 48

¡

and S6 ¼ 0:606 510 42

0 < bn+1 < bn for all n 2 Z + ) fbn g is a decreasing sequence, n 2 Z + .

1

1 8

+

But, by the Estimation Theorem jS ¡ S6 j 6 b7 ) S6 ¡ b7 6 S 6 S6 + b7 ) 0:606 489 6 S 6 0:606 532 ) S ¼ 0:6065

n=1

Also

1 2

1 46 084

where b7 =

So, by the Estimation theorem jS ¡ S5 j 6 b6 ) ¡0:001 39 6 S ¡ 0:633 33 6 0:001 39 ) 0:631 94 6 S 6 0:634 72 Thus S ¼ 0:63

Now as 0
1

Thus

1

n2 is convergent fas a1 = 12 g n3 + 1

0 < bn+1 6 bn for all n 2 Z + Also

1 (¡1)n is an alternating series with bn = n 2n n! 2 n!

¡ Now bn+1 ¡ bn = n+1 2 (n + 1)! 2n n!

95

a

(¡1)n+1

S ¼ 0:8415

1 P

100

4

1 P

1 n3 n2 is convergent n3 + 1

1+

n=2 1 P

) =0

black



WORKED SOLUTIONS Thus S4 ¡ S2 = b1 ¡ b2 + b3 ¡ b4 ¡ (b1 ¡ b2 ) = b3 ¡ b4 which is > 0 fas fbn g is decreasingg

¼

1 P

But

n=1

Thus S2n > S2n¡2 > S2n¡4 > :::: > S4 > S2 So 0 6 S2 6 S4 6 :::: 6 S2n¡4 6 S2n¡2 6 S2n

)

c S2n = b1 ¡ b2 + b3 ¡ b4 + :::: ¡ b2n¡2 + b2n¡1 ¡ b2n ) S2n = b1 ¡ (b2 ¡ b3 ) ¡ (b4 ¡ b5 ) ¡ :::: ¡ (b2n¡2 ¡ b2n¡1 ) ¡ b2n

1 P n=1

)

1 P

)

1 P n=1

S2n 6 b1

d From b and c, fS2n g is increasing and has an upper bound of b1 lim S2n = S.

n!1

= b1 ¡ b2 + b3 ¡ b4 + :::: + b2n+1 ¡ b2n + b2n+1

}

)

S2n S2n+1 = S2n + b2n+1

)

n!1

n!1

n!1

=S

a If an =

)

n!

,

¯ ¯ ¯ ¯ 3n+1 an+1 ¯ ¯ lim ¯ = lim ¯¯ ¯ n!1 n!1

¯ n! ¯¯ (n + 1)! 3n ¯

an

= lim

n!1

¯ ¯ ¯ an+1 ¯ ¯ ¯ n!1

(¡1)n 2n , n2 + 1

b If an =

lim

2n+1 = lim n!1 (n + 1)2 + 1 = lim 2 n!1

0

n!1

=2£

cyan

1 + n12 1+

magenta

2 n

+

2 n2

4

fa convergent GSg fComparison testg

is absolutely convergent.

ln x x

1 x x

¡ ln x

=

x2

1 ¡ ln x x2

f 0 (x) < 0 for all x such that ln x > 1, that is, x > e

n ln n o n

is decreasing for all n > 3, n 2 Z + .... (1) 1

lim

1 P n=1

n2 + 1 2n

¶

f Let bn =

1 , then n ln n

Now consider f (x) =

¶

(¡1)n+1 ln n n

is absolutely

lim bn = 0

n!1

1 = [x ln x]¡1 , x > 2 x ln x

³

) f 0 (x) = ¡[x ln x]¡2 1 ln x + x

1

¡(ln x + 1) = (x ln x)2

A

³ 1 ´´ x -

2

So, f 0 (x) < 0 for all x > 2 1 is decreasing for all n > 2, n 2 Z + ) bn = n ln n Thus

1 P

yellow

95

50

75

n=2

95

50

75

25

0

4n + 3

¡ 3 ¢n

ln x = lim x = 0 x!1 x n!1 1 ln n ) lim = 0, n 2 Z + .... (2) n!1 n

1 1

5

95

100

50

75

25

0

95

100

50

75

25

0

5

n=1

3 + 4n

By l’Hôpital’s Rule,

=2 (¡1)n 2n is divergent fRatio testg n2 + 1

5

)

1 P

µ

n2 + 1 n2 + 2n + 2

= lim 2@

jan j is convergent

³ 3n ¡ 1 ń

convergent. fAlternating Series Testg

an

µ

jan j
0, b4 ¡b5 > 0, ...., b2n¡2 ¡b2n¡1 > 0, and b2n > 0

e

jarctan nj 2 6 3 n3 n

jan j =

and as S2n ¡ S2n¡2 = b2n¡1 ¡ b2n > 0 S2n ¡ S2n¡2 > 0 for all n 2 Z + .

)

(¡1)n arctan n then n3

c If an =

Likewise S6 ¡ S4 > 0, S8 ¡ S6 > 0, etc,

)

179

black



180

WORKED SOLUTIONS g Let br = ) br =

p p r+1¡ r

¡p

r+1¡

p

r

¢

p ¶ r+1+ r p p r+1+ r

µp

r+1¡r p r+1+ r 1 1 ) br = p p > p r+1+ r 2 r+1 1 P r=1 1

P

)

br >

1 2

br >

1 2

r=1

P 1

where

n=2 1 P

)

1 P

Hence

1 P

c

lim

n!1

1 P

)

fn = r + 1g

2n 2 = lim 8n ¡ 5 n!1 8 ¡

n=1

1 p is divergent fp-series testg n

d Let an =

cos

p p (¡1)r¡1 ( r + 1 ¡ r) is divergent.

xn for x 2 R . n!

a Let an =

¯ ¯ ¯ ¯ an+1 ¯ = ¯¯ xn+1 ¯ a ¯ ¯ (n + 1)! n ¯ ¯ ¯ an+1 ¯ = 0 lim ¯ ¯ n!1 a

¯

)

jan j 6

)

06

)

n=1

)

)

n=0

)

xn is absolutely convergent and ) convergent. n!

an is convergent )

n=0

e

lim an = 0

n!1

n3 n4

1 P

9

a

1 P n=0

)

10n is n!

1 P n=0 1 P

by 8 a,

n=0

1

(n + 1)(n + 1)

1,

n=1

1 = n+1

n=1

µ

P 1

n=1

1 n

2

is absolutely convergent and

n3 +1 > 4 for all n 2 Z + , n > 2 ¡1 n 1 P n3 + 1 1 > n4 ¡ 1 n n=2

1 P 1 P

1 diverges n

fp-series testg

n3 + 1 diverges n4 ¡ 1

fComparison testg

n! 2 £ 5 £ 8 £ :::: £ (3n + 2) ¯a ¯ (n + 1)! ¯ n+1 ¯ = ) ¯ ¯ an 2 £ 5 £ 8 £ :::: £ (3n + 5) 2 £ 5 £ 8 £ :::: £ (3n + 2) £ n! n+1 = 3n + 5 Let an =

)

¶

fComparison testg

convergent.

n=2

xn with x = 10 n!

1 n2

¡n¢

n2 + 4n

n=2

)

1 P n=1

cos

n=2

But

and n2 + 4n > n2 , n 2 Z +

jan j is convergent

1 P

)

xn is convergent n=0 n! xn =0 ) lim n!1 n!

)

jan j 6

1 P

)

b By the Converse of the Test for divergence 1 P

2

1 is convergent fp-series testg n2

n=1

n

1 P

1 P

6= 0

1 n2

1 P

1 P

where

n! ¯¯ jxj = xn ¯ n+1

1 4

n2 + 4n

n=1

Now

=

¡n¢

¯ ³ ´¯ ¯cos n ¯ 6 1 ¯ ¯ 2

Now

br is divergent

5 n

2n is divergent. 8n ¡ 5

r=1

8

diverges

n(n + 1)

fComparison test for seriesg

r=1

)

1

p

n=1

1 p r+1 r=1 p n n=2

diverges fp-series testg, it follows that

1 diverges. n+1

n=1

1 P

1 P 1

1 n

n=1

1 P

) br = p

)

1 P

Since

)

¯ ¯ ¯ an+1 ¯ = lim ¯ ¯ n!1 n!1 a lim

n

by the Ratio test

¡ 1, which is the

1 P

1+ 3+

1 n 5 n

=

1 3

6= 0

an converges.

n=0

cyan

magenta

yellow

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

Harmonic Series with the first term removed.

black



181

WORKED SOLUTIONS 1 , n2

10 If an =

¯ ¯ ¯ an+1 ¯ = lim ¯ ¯ n!1 n!1 lim

1 n2 (n + 1)2 1

an

n2 n2 + 2n + 1 1

= lim

n!1

= lim

by the Ratio test

1 1 1 1 = 1 + 2 + 4 + 6 + :::: is geometric with a1 = 1 2n x x x x n=0 1 and r = 2 . x )

1 n2

1 P

n=0

=1

1 P n=1

divergent.

1 P

2 1+ n + n12

n!1

)

2

But, from the p-series test it is convergent. Likewise, if an =

1 , n

¯ ¯ ¯ an+1 ¯ = lim ¯ ¯ n!1 n!1 lim

an

= lim

n!1

)

¯ 2¯ ¯x ¯ > 1

)

jxj > 1

)

1 n n+1 1

)

1

3

x2 x2

¶

1 x2 ¯ ¯ ¯ 1 ¯ 1 or x < ¡1

the interval of convergence is ]¡1, ¡1[ [ ]1, 1[ .

a If an = n5n xn ,

¯ ¯ ¯a ¯ ¯ (n + 1)5n+1 xn+1 ¯ ¯ n+1 ¯ ¯ ¯ lim ¯ lim ¯ ¯ = n!1 ¯ n!1 an n5n xn ¯³ ¯ ´ 1 ¯ ¯ = lim ¯ 1 + 5x¯ n!1

1 1+ n

=1 So, once again the ratio test is inconclusive.

n

= j5xj

EXERCISE K.4 1

a

1 P

= 5 jxj

x3n = 1+x3 +x6 +x9 +x12 +:::: which is geometric

with u1 = 1 and r = 1 P

)

x3n =

n=0

x3 .

b

(2 ¡

x)n

1 u1 = 1¡r 1 ¡ x3

provided

and diverges for jxj >

jrj < 1

)

¯ 3¯ ¯x ¯ < 1

)

jxj < 1

)

= 1 + (2 ¡ x) + (2 ¡

x)2

+ (2 ¡

x)3

each case (x = If x =

1 , 5

+ ::::

P

1 P

j2 ¡ xj < 1 jx ¡ 2j < 1 ¡1 < x ¡ 2 < 1 1<x 0, [x ln x]2 > 0g

= j2x + 3j

25

1 P

1 is divergent. n ln n

n!1

n!1

0

1 P

(ln x + 1) =¡ [x ln x]2

(¡1)n (2x + 3)n d If an = , ¯n ln n ¯

5

ln 2

) f 0 (x) = ¡[x ln x]¡2 1 ln x + x

It has an infinite radius of convergence and its interval of convergence is R .

cyan

when x = 2, u = ln 2 when x = b, u = ln b

¤ ln b ´

by the Integral test,

n > 2.

an is convergent for all x 2 R .

fas

ln juj

For x = ¡1, we have

n=0

n!1

ln 2

1 du u

n=2

, then

= lim j2x + 3j £

¶ ½

ln b

du 1 = g dx x

b!1

=0

1 P

³£

fu = ln x,

= lim (ln(ln b) ¡ ln(ln 2))

¯ ¯ ¯ ¯ ¯ an+1 ¯ = lim ¯¯ x2n+1 (2n ¡ 1)! ¯¯ lim ¯ ¯ n!1 ¯ n!1 an (2n + 1)! x2n¡1 ¯ ¯ ¯ ¯ ¯ x2 ¯ ¯ = lim ¯ n!1 2n(2n + 1) ¯ )

¶

which DNE

is convergent for ¡ 13 6 x 6

It has a radius of convergence of

1 du dx u dx

100

)

3n xn

b

2

µZ

b!1

(¡1)n (n + 1)2

which converges by the Alternating series test. 1 P

1 dx x ln x

2

µZ

¶

b

= lim

1 converges. (n + 1)2

3n ¡ 13

µZ

= lim

1 converges fp-series testg n2

1 P

n=2

1 n ln n

1 dx x ln x

= lim

1 1 < where (n + 1)2 n2 n=0

1 P

1

Now

1 P

n=0

If x = ¡ 13 ,

1 , 3

1 P

=

1)2

(n +

by the Comparison test

1 P

If x = ¡2, we have

fRatio testg

As the Ratio test is inconclusive for jxj =

)

¡1 < 2x + 3 < 1 ¡4 < 2x < ¡2

j3xj < 1 ) jxj
1 also

)

Thus

P

and

£ (¡1)n

n=1

R

n=0 1

lim an 6= 0

P

p

p R.

n=0

n!1 1

)

1 P

7

1 P

where

jan j > 1 fas each fraction is > 1g

)

)

¡8¢

1 P

(cn + dn )xn =

n=0

2 £ 4 £ 6 £ 8 £ :::: £ (2n)(¡1)n 1 £ 3 £ 5 £ 7 £ :::: £ (2n ¡ 1)

¡4¢

the radius of convergence is

1 P

7

As the Ratio test is inconclusive for x = §1, we examine each of these cases separately.

=

cn x2n converges when jxj
1 or x < ¡1 ) x 2 ]¡1, ¡1[ [ ]1, 1[

=0 xn¡1

1 P

£ x2n ¯ =

x2(n+1)

) interval of convergence

x

¶

or

(¡1)n x2n

jxj = lim n!1 n 1 P

0

(0:1)n ¡0 n

1 n n(n ¡ 1)! 10 n=1

n

¶

0:1

1 P

¯ ¯ ¯ ¯ an+1 ¯ = ¯ ¯ a ¯ ¯

n=1 (n ¡ 1)!

¯ ¯ ¯ ¯ ¯ an+1 ¯ = lim ¯¯ xn (n ¡ 1)! ¯¯ ¯ ¯ n¡1 ¯ n!1 n!1 ¯ lim

where

1 P

dx

(n ¡ 1)!

1 = (n ¡ 1)! n=1

dx n!

n=1

n=1

1 P

´ 1 ³ P d xn

=

¶

1 P xn¡1

1 = (n ¡ 1)! n=1

nxn¡1 converges for all x 2 ]¡3, 3], and has radius n2 3n

1 P xn

xn¡1 dx is defined (n ¡ 1)!

1 P

of convergence 3.

µ

µ

0

converges conditionally.

n=0

0:1

)

(¡1)n¡1 When x = ¡3, an = and 3n 1 1 P P (¡1)n¡1 an = which 3n n=1 n=1 1 P

n=1

Z

divergent. fp-series testg

)

P

0:1 1

)

When x = ¡1,

1 P

(¡1)2n =

n=0

by Example 38 a this has an

since

1 P

n=0

1n =

1 P

1 diverges,

n=0

lim 1 6= 0.

n!1

cyan

magenta

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5

95

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5

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5

infinite radius of convergence.

black



WORKED SOLUTIONS 1 P

When x = 1,

1 P

12n =

n=0

1 diverges,

n=0

lim 1 6= 0.

since

n!1

)

the interval of convergence is ]¡1, 1[

Z

1 2

µ

0

¶

1 P

x2n

1 P

dx =

n=0

ÃZ

1 2

2n + 1

n=0 1 P

1 = n=0 2n + 1

Z 2µ

) 0

1 P

0

³¡ ¢2n+1 1

)

0:5805 < sin

¡0

2

1 (2n + 1) £ 22n+1 n=0

¡ sin

5

¡¼¢

¡ ¼ ¢5 5

¡¼¢

¡ ¼5¢ 5

< 0:0065

< 0:0065

< ¡0:5805

< 0:5935

the approximation is accurate to 1 decimal place only. x5 x3 + + R5 (x : 0) 3! 5!

¯ ¯ ¯ f (6) (c)x6 ¯ ¯ ¯ jR5 (x : 0)j = ¯ ¯ 6! ¯ ¯ ¯ ¡ sin c £ x6 ¯ ¯ ¯ =¯ ¯ 6! =

x2n has an interval of convergence

jsin cj jxj6 6!

and c lies between 0 and x. But jxj 6 0:3 and max jsin cj = sin(0:3)

dx is not defined as the interval [0, 2]

x

¡0:5935 < ¡ sin

where

´

¶ 2n

)

2 sin x ¼ x ¡

n=0

of ]¡1, 1[ .

¡0:0065 < 0:5870 ¡ sin

1 P

1 P

11 From question 10 part c,

)

)

x2n dx

0 1 · 2n+1 ¸ 12 1 P x @ A

=

¡0:0065 < T3

0

n=0

=

!

¡¼¢

)

185

n=0

sin(0:3)(0:3)6 6!

)

jR5 (x : 0)j
0 for all k, 1 + uk 6 euk (1 + u1 )(1 + u2 )(1 + u3 ) :::: (1 + un ) 6 eu1 eu2 eu3 ::::eun

yellow

95

´³

´

x x x2 1+ =1¡ 2 ¼ ¼ ¼ ³ ´³ ´ x x x2 1¡ 1+ =1¡ 2¼ 2¼ 4¼ 2 ³ ´³ ´ x x x2 1¡ 1+ =1¡ , and so on. 3¼ 3¼ 9¼ 2 1¡

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

= eu1 +u2 +u3 +::::+un

5

x2 (2n + 3)(2n + 2)

c The zeros of ³ ´³ ´³ ´³ ´³ ´ x x x x x 1¡ 1+ 1¡ 1+ 1¡ :::: ¼ ¼ 2¼ 2¼ 3¼ are §¼, §2¼, §3¼, §4¼, ....

) ex ¡ (1 + x) > 0 for all x > 0

magenta

¯ ¯ ¯ an+1 ¯ ¯ ¯ n!1 an ¯ ¯ ¯ x2n+2 (2n + 1)! ¯ ¯ ¯ = lim ¯ n!1 (2n + 3)! x2n ¯ lim

n!1

x3 x4 x2 + + + :::: 2! 3! 4! x2 x3 x4 Now for x > 0, + + + :::: is > 0 2! 3! 4!

cyan

(¡1)n x2n , (2n + 1)!

¸

µ3 µ5 µ7 + ¡ + :::: 3! 5! 7!

a ex = 1 + x +

)

(2n + 1)!

= lim

b eiµ = cos µ + i sin µ 6

1 P (¡1)n x2n

¸

·

(1, 0)

)

iµ i2 µ2 i3 µ3 i4 µ 4 i5 µ 5 + + + + + :::: 1! 2! 3! 4! 5!

+i µ¡

(-1, 0)

sin x = 0 , x = k¼, k 2 Z , x 6= 0 x sin x = 0 , x = §¼, § 2¼, § 3¼, .... x x3 x5 x7 sin x = x ¡ + ¡ + :::: 3! 5! 7!

b

k!

= 1¡

y

x

1 (iµ)k P k=0

(1 + uk ) has an upper bound of eL

(1 + uk ) is increasing and has an upper bound, it

zk for all z k!

k=0

(1 + uk ) 6 eL

k=1

¶

a0 = 0, a1 = 1, a2 = 0, a3 = tan x ¼ x +

un converges

n=1

k=1

By equating coefficients we get

)

Q

As

= a0 + a1 x + (a2 ¡ 12 a0 )x2 + (a3 ¡ 12 a1 )x3 + (a4 ¡ 12 a2 +

n Q

1 P

k=1 n

sin x , sin x = cos x tan x cos x

as tan x =

n Q

Thus

¡ :::: and

5!

(1 + uk ) is increasing and as

k=1

x2 x4 cos x = 1 ¡ + ¡ :::: 2! 4! Let tan x = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + :::: )

n Q

)

¼ That is, for x 2 R , x 6= (2k + 1) , k 2 Z . 2 x5

(1 + uk ) 6 eu1 +u2 +u3 +::::+un

k=1

b cos x has R = 1 and is defined for all x 2 R . x equals its full Maclaurin series expansion for all ) cos x x is defined. x 2 R such that cos x

x3

n Q

)

black



193

WORKED SOLUTIONS

³

)

x ¼

1¡

µ

´³

x ¼

1+

x2 ¼2

1¡

=

´³

¶µ

1¡

1¡

x2 4¼ 2

x 2¼

´³

¶µ

1¡

x2 9¼ 2

´³

x 2¼

1+

1¡

¶

x 3¼

´

y = 2x ¡ 2 + ce¡x is a solution of

)

::::

b We need graphs of y y y y y

::::

sin x has zeros of §¼, §2¼, §3¼, .... x ³ ´³ ´³ ´³ ´ x x x x From c, 1 ¡ 1+ 1¡ 1+ :::: has ¼ ¼ 2¼ 2¼ From a,

= 2x ¡ 2 = 2x ¡ 2 + e¡x = 2x ¡ 2 ¡ e¡x = 2x ¡ 2 + 2e¡x = 2x ¡ 2 ¡ 2e¡x

c=2

the same zeros.

dy = 2x ¡ y dx (c = 0) (c = 1) (c = ¡1) (c = 2) (c = ¡2)

y 4

This evidence supports Euler’s claim that

sin x x2 x4 x6 =1¡ + ¡ + :::: is equal to x 3! 5! 7!

µ

x2 1¡ 2 ¼

e Now,

µ

x2 1¡ 2 ¼

µ

x2 1¡ 4¼ 2

¶µ

x2 1¡ 4¼2

¶µ 2

x ¼2

1¡

and

¶µ

h

1¡

³1 2

1 + ¼2 4¼ 2

= 1¡x =1 ¡ x2

x2 4¼2

³1

¼2

+

¶µ

2

=1¡x

¶µ 1¡

´

¶

x2 1¡ 9¼ 2

¶

:::: (1)

³1

´

c=0

¶

x2 1¡ 9¼ 2

¶

1 9

1 16

+

+ ::::

That is, y = ¡x + 1

+ :::: 2

1 P 1 1 = = 2 2 (2n) 4n n=1 n=1 1 P

1 ) = 2 n=1 (2n)

n=1

n2

=

µ

1 4

¼2 6

¶

p

x2 + c then y 2 = x2 + c

1 P 1 n=1

)

n2

2r

+

p x dy = . x2 + c is a general solution of dx y p b As y(3) = 4, 4 = 9 + c ) 16 = 9 + c ) c=7 p Thus y = x2 + 7

1 P r=1

1 , (2r ¡ 1)2

1 P

1 ¼2 ¼2 ¼2 = ¡ = 2 6 24 8 n=1 (2n ¡ 1)

3

a

y 2 1

EXERCISE M 1

a If y = 2x ¡ 2 + ce¡x then -2

dy = 2 + ce¡x (¡1) dx

magenta

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2 x

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5

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0

5

95

1

-2

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5

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25

0

cyan

-1 -1

= 2 ¡ ce¡x = 2 ¡ [y ¡ 2x + 2] = 2 ¡ y + 2x ¡ 2 = 2x ¡ y

5

dy = 2x dx dy x = dx y

Thus y =

¼2 = 24

´ 1 ³ P 1 2 r=1

1 4

1 P ¼2 ¼2 1 = + 6 24 n=1 (2n ¡ 1)2

)

a If y =

) 2y

1 P

1 P 1

y¡1 = ¡1 x¡0

) the equation is

¼2 ) = 2 6 n=1 n

and as

dy = 2(0) ¡ (1) = ¡1 dx

d At (0, 1),

1 P 1

f

c = -2

c If (0, 1) lies on y = 2x ¡ 2 + ce¡x then 1 = ¡2 + c ) c = 3 ) y = 2x ¡ 2 + 3e¡x

´

+

x

-4

1 1 1 =¡ + + + :::: 3! ¼2 4¼2 9¼ 2 1 4

4

c = -1

³1

¼2 =1+ 6

)

2 -2

By equating the coefficients of x2 in (1) we get ¡

-2

´

iµ

1 1 + 4¼ 2 9¼2

-4

1 + + :::: ¼2 4¼ 2

x2 9¼ 2

+ ::::

2

c=1

black

Y:\HAESE\IB_HL_OPT-Calculus\IB_HL_OPT-Calculus_AN\193IB_HL_OPT-Calculus_an.cdr Wednesday, 13 March 2013 2:43:29 PM BRIAN


194

WORKED SOLUTIONS b

6

y

a

y

2

2

1

1 -3

x -1

-2

1

-2

-1 1

2

2

x

3

-1 -1 -2 -2

y

¡2 ¡1 0 1 2

¡2 0:70 0:35 0 ¡0:35 ¡0:70

dy is undefined when y = 5x ¡ 10 dx dy = 0 when x2 + 4y 2 = 1 ii dx

b

dy 4 = 10y tan x, x in degrees dx

x 0 0 0 0 0 0

¡1 0:35 0:17 0 ¡0:17 ¡0:35

1 ¡0:35 ¡0:17 0 0:17 0:35

i

2 ¡0:70 ¡0:35 0 0:35 0:70

y

y = 5x - 10

2 1 -3

-2

-1

x2 + 4y2 = 1

y 2

1

2

x

3

-1 -2

1

-1

2 x

1

nx

n+1 = xn + h yn+1 = yn + hf (xn , yn ) where h = 0:2

and f (xn , yn ) = 1 + 2xn ¡ 3yn

-1

) yn+1 = yn + 0:2(1 + 2xn ¡ 3yn ) = 0:2 + 0:4xn + 0:4yn

-2

2

x0 x1 x2 x3 x4 x5

2

5 If k = 0,

x + y ¡ 1 = 0 ) y = ¡x + 1

If k = 2, If k = 4,

x2 + y ¡ 1 = 2 ) y = ¡x2 + 3 x2 + y ¡ 1 = 4 ) y = ¡x2 + 5 y

8 -4

n+1 = xn + 0:1 yn+1 = yn + 0:1(sin(xn + yn ))

x0 x1 x2 x3 x4 x5

x y = -x2 + 5 -4 y = -x2 + 3

y = -x2 + 1

-2

=0 = 0:1 = 0:2 = 0:3 = 0:4 = 0:5

dy = 1, y(4) = 3 dx dy 1 ) = dx 2¡x Z 1 ) y=¡ dx x¡2

a (2 ¡ x)

1

1 1

2

4 x

3

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yellow

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5

95

100

50

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25

0

5

95

100

50

) y = ¡ ln jx ¡ 2j + c

75

25

0

¼ 0:5 ¼ 0:547 94 ¼ 0:608 30 ¼ 0:680 61 ¼ 0:763 69 ¼ 0:855 52

EXERCISE N

-1-1

5

95

100

50

75

25

0

5

y0 y1 y2 y3 y4 y5

So, y(0:5) ¼ y5 ¼ 0:856

y

-2 -3 -4 -5

cyan

=1 = 0:6 = 0:52 = 0:568 = 0:6672 = 0:786 88

nx

4

-4 -3

y0 y1 y2 y3 y4 y5

So, y(1) ¼ y5 ¼ 0:787

4

5 4 3 2

=0 = 0:2 = 0:4 = 0:6 = 0:8 =1

100

-2

7

black



WORKED SOLUTIONS

Z

But when x = 4, y = 3 ) 3 = ¡ ln 2 + c ) c = 3 + ln 2 Hence, y = 3 + ln 2 ¡ ln jx ¡ 2j

But, when x = e, y =

x¡2

dy ¡ 3x sec y = 0, y(1) = 0 dx dy ) = 3x sec y dx dy cos y = 3x dx Z Z dy ) cos y dx = 3x dx dx

Z

)

2

a

3 2

) c= Hence, sin y =

2

3 2

¡

£3

) y = arcsin

2

=

3 (x2 2

)

+c

Z

¡ 32

)

¡ 1)

¤

(x2 ¡ 1)

Z

1 4

ey dy = y e +3

1 4

4x + 4 dx 2x2 + 4x + 1

) ln jey + 3j =

1 4

ln ¯2x2 + 4x + 1¯ + c

Z

Thus,

4x + 4 dx + 4x + 1

ey dy dx = y e + 3 dx

)

¯

)

¯

¯ ¯1 + 3) = ln ¯2x2 + 4x + 1¯ 4

) ey =

+c

d

Hence y =

µ ¯1 ¯ 2 ¯2x + 4x + 1¯ 4

Z

¶

3

(e2 + 3)

j2x2 + 4x + 1j (e2 + 3) ¡ 3

hp 4

a

(x ¡ 1) then y = ¡1 (x + 1)2 ¡(x ¡ 1) then y = 1 (x + 1)2

1¡x (x + 1)2

dT / T ¡ R, t > 0 dt

dT = k(T ¡ R) dt 1 dT ) =k T ¡ R dt Z Z dT 1 ) dt = k dt T ¡ R dt

i

j2x2 + 4x + 1j (e2 + 3) ¡ 3

dy = cos2 y, y(e) = dx dy 1 ) sec2 y = dx x sec2 y

(x ¡ 1) (x + 1)2

From a,

x

)

´

2 dx x+1

jx ¡ 1j (x + 1)2

which is true.

p 4

) y = ln

´

¯ ¯ ¯ x¡1 ¯ ¯ (x + 1)2 ¯

When x = 0 and y =

c = ln(e2 + 3)

Thus, ln(ey + 3) = ln

x¡1

¡

which is false.

ln(e2 + 3) = ln 1 + c

)

Z ³ 1

Check: When x = 0 and y =

But when x = 0, y = 2 )

³

ln jyj = ln ¯¯

) y=§

But ey + 3 > 0 for all y 2 R ln(ey

3¡x x2 ¡ 1

1 dy dx = y dx

) jyj =

2x2

Z

=

But when x = 0, y = 1 ) 0 = 0 ¡ 2(0) + c ) c=0

dy c e (2x + 4x + 1) = (x + 1)(ey + 3), y(0) = 2 dx ³ ey ´ dy x+1 ) = ey + 3 dx 2x2 + 4x + 1

Z

x + 1 ¡ 2x + 2 (x ¡ 1)(x + 1)

) ln jyj = ln jx ¡ 1j ¡ 2 ln jx + 1j + c

2

)

=

dy 3y ¡ xy 3¡x = 2 =y dx x ¡1 x2 ¡ 1 1 dy 1 2 = ¡ ffrom ag y dx x¡1 x+1

b

But when x = 1, y = 0 ) 0 =

y

1 2 (x + 1) ¡ 2(x ¡ 1) ¡ = x¡1 x+1 (x ¡ 1)(x + 1)

3x dx

3x2 ) sin y = +c 2

3x2

) 1=1+c ) c=0

¼ 4

) tan y = ln jxj ) y = arctan(ln jxj)

Z

cos y dy =

1 dx x

) tan y = ln jxj + c

¯ ¯ ¯ 2 ¯+3 y = ln ¯ ¯

b

Z sec2 y dy =

)

195

dy dx = dx

Z

¼ 4

Z

)

1 dT = T ¡R

Z

k dt

) ln jT ¡ Rj = kt + c

1 dx x

) T ¡ R = §ekt+c ) T = §ec ekt + R

cyan

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yellow

95

100

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25

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5

95

100

50

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25

0

5

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5

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5

) T = Aekt + R .... (1)

black



196

WORKED SOLUTIONS b But when t = 0, R = 18, and T = 82

5

AP : PB = 2 : 1 ) A is (3x, 0) and B is (0, 32 y)

y

) 82 ¡ 18 = Ae0 ) A = 64 Thus (1) becomes T = 64ekt + 18 .... (2)

B

But when t = 6, T = 50 ) 50 = 64e6k + 18

P(x, y)

1

2

) 64e6k = 32 )

e6k =

)

ek =

¡ 1 ¢ 16 2

¡ 1 ¢ 6t

Thus (2) becomes, T = 64

2

+ 18

t ¡6

or T = 64 £ 2

+ 18

Z

t

¡6

Now when T = 26,

26 = 64 £ 2

+ 18

=8

) 64 £ 2

t

¡6

)

2

)

=

1 8

=2

+ 18

) 64 £ 2

=2

t

)

2

¡6

)

=

1 32

1 x

) y = §p

= 2¡5

So, from t = 18 to t = 30, T decreases from

26±

to

20± .

This takes place over a 12 minute time interval.

y x gradient of tangent at x P(x, y) is ¡ y x dy =¡ dx y

Gradient of [OP] = ) P(x, y)

)

6

a

Z (0, 0)

Hence, y

Z )

)

x

Q

) ln jmj = kt + c ) jmj = ekt+c ) m = §ec ekt

y dy = ¡

x dx

) m = Aekt .... (1) But at t = 0, m = m0 ) m0 = A

Z

y2 2

=¡

x2 2

)

b At t = 30, m(t) = ) (2) becomes

But when x = 1, y = 2 ) 1 + 4 = 2c ) 2c = 5

magenta

yellow

=

95

4 5

1 ¡ 4 ¢ 30

5

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

= m0 e30k

) ek =

5

95

100 cyan

.... (2)

4 m 5 0

4 m 5 0 30k

) e

x2 + y2 = 5

75

50

(1) becomes m = m0 ekt

That is, m(t) = m0 ekt

+c

) x2 + y 2 = 2c

25

k dt

x dx

)

0

Z

dy dx = ¡ dx

)

5

1 dm dt = m dt

dy = ¡x dx

Z

)

fas x must be > 0g

dm /m dt dm ) = km dt 1 dm ) =k m dt

Z

y

1

¡2

1 Check: When x = 1 and y = p , y = 1 X x 1 When x = 1 and y = ¡ p , y = ¡1 x 1 ) y= p x

t = 30

y

1 dx x

Thus, ln jyj = ¡ 12 ln jxj = ln jxj

t

¡6

Z

But when x = 1, y = 1 ) 0=0+c ) c=0

t

¡6

20 = 64 £ 2

1 dy dx = ¡ 12 y dx

) ln jyj = ¡ 12 ln jxj + c

¡3

t = 18

and when T = 20,

¡ 32 y dy OB ¡y ) =¡ = = dx OA 3x 2x 1 dy ¡1 ) = y dx 2x

)

t

¡6

4

x

A

1 2

black



WORKED SOLUTIONS Hence in (2), m(t) = m0

¡ 4 ¢ 30t

Since 1 + v2 > 0 for all v,

5

m(t) =

Now when

1

arctan v = ln(1 + v2 ) 2 + ln jxj + c

1 m , 2 0

)

(0:8)

) arctan

= 0:5

t ln(0:5) = 30 ln(0:8)

) )

c

x¡y dy = a dx x Let y = vx,

dv Hence x +v dx dv ) x dx 1 dv ) 1 ¡ 2v dx Z 1 ) dv 1 ¡ 2v

dy y2 ¡ x2 = dx 2xy Let y = vx

dv dy = x+v dx dx

)

) x

1 dx x

)

Z

1 ln j1 ¡ 2vj = ln jxj + c ¡2 ¯ ¯ 2y ¯ ¯ ) ln ¯1 ¡ ¯ = ¡2 ln jxj ¡ 2c x ¯ ¯ ¯ x ¡ 2y ¯ + ln ¯¯x2 ¯¯ = ¡2c ) ln ¯ ¯ x ¯ ¯ ¯ x ¡ 2y £ x2 ¯ = ¡2c ) ln ¯ ¯ x

Z

2v dv dx = ¡ 1 + v 2 dx

)

)

¯

¯µ ¶ ¯ ¯ ¯ y2 ¯ ln ¯ 1 + 2 x¯¯ = c x

)

) x+

8

) x2 ¡ 2xy = A, a constant

) x

dy dv = x+v dx dx dv x + vx 1+v +v = = ) x dx x ¡ vx 1¡v dv 1+v ) x = ¡v dx 1¡v dv 1 + v ¡ v(1 ¡ v) 1 + v2 = = ) x dx 1¡v 1¡v )

)

cyan

y dy = y + e x then dx y³ ´ dy y 1 = + ex dx x x 1 dv 1 So, by a, = 2 ev dx x

¯

¯

Z

)

1 dx x

magenta

yellow

95

1 dv = ev 1 ¡v e ¡1

=

Z

x¡2 dx

x¡1 +c ¡1

100

50

75

25

0

)

5

95

100

50

ln ¯1 + v 2 ¯ = ln jxj + c

75

1 2

25

95

50

75

25

0

5

95

100

50

75

25

0

5

) arctan v ¡

2v dv = 1 + v2

0

1 2

5

1 dv ¡ 1 + v2

100

)

Z

dv = f (v)g(x) dx 1 dv g(x) = and so is separable. f (v) dx x

b If x

1 ¡ v dv 1 = 1 + v2 dx x Z ³ Z ´ dv 1 v 1 ) ¡ dx = dx 1 + v2 1 + v 2 dx x

Z

dv + v = v + f (v)g(x) dx

) x

Thus,

Z

dv dy = x+v dx dx ³y´ dy y = +f g(x) becomes dx x x

a If y = vx,

x+y dy = dx x¡y Let y = vx

y2 = §ec = A, say x

) x2 + y2 = Ax, where A is a constant

¯ 2 ¯ ¯x ¡ 2xy¯ = e¡2c

) x2 ¡ 2xy = §e¡2c

b

¯

1 dx x

) ln ¯1 + v2 ¯ = ¡ ln jxj + c

) ln jx(x ¡ 2y)j = ¡2c )

dv v2 ¡ 1 = ¡v dx 2v

dv v 2 ¡ 1 ¡ 2v 2 ¡v 2 ¡ 1 = = dx 2v 2v 2v dv ¡1 = 1 + v 2 dx x

1 x

Z

dy dv = x+v dx dx

dv v 2 x2 ¡ x2 v2 ¡ 1 +v = = dx 2x(vx) 2v

) x

= 1 ¡ 2v

=

)

) x

y =1¡ =1¡v x

=

x2 + y 2 + c

= ln

t ¼ 93:2

It would take 93:2 days (approximately). 7

¯ r ¯ ¯ ¯ y2 ¯ ¯ = ln ¯x 1 + 2 ¯ + c x x ¯ ¯ ¯p ¯ ¯ ¯ = ln ¯ x2 + y 2 ¯ + c p

³y´

t

m0 (0:8) 30 = 0:5m0 t 30

197

black



198

WORKED SOLUTIONS y

¡x

) ¡e

=¡

y

¡x

) e

=

h

Z

1 +c x

1 ¡c x

) y = x ¡ 1 + 12 ex + ce¡x

i

But when x = 1, y = 1 ) 1 = 1 ¡ 1 + 12 e + ) 1¡

h 1 ¡ cx i¡1

) y = x ¡ 1 + 12 ex + e1¡x ¡ 12 e2¡x d x dy + P (x)y = Q(x) dx

dy + 4y = 12 is of the form dx with P (x) = 4

R

)

4 dx

the integrating factor is e

Multiplying the DE by

e4x

= e4x .

= 12

)

¡1¢ 4

) y = 3 + ce

4x

R

Z

=

) xy = x sin x ¡

) y=

) y = sin x +

¡ 12

+c ) c=

1 5 Thus y = ¡ ex + e3x . 2 2 c

dy + y = x + ex dx

R

has I(x) = e

)

5 2

= ex

We integrate by parts with:

Z e2x dx

) xex y = ex (x ¡ x2 ) ¡

magenta

= ex+ln x = ex £ x

yellow

u0 = ex u = ex

95

v = x ¡ x2 v 0 = 1 ¡ 2x

ex (1 ¡ 2x) dx

n

u0 = ex u = ex

100

50

25

0

5

95

100

50

75

25

0

R

ex (x ¡ x2 ) dx

) xex y = ex (x ¡ x2 ) ¡ [ex (1 ¡ 2x) ¡

5

95

v=x v0 = 1

100

50

We integrate by parts with:

75

xex dx +

75

25

0

5

95

100

50

75

25

0

5

cyan

1+ x dx

n

(xex + e2x ) dx

u0 = ex u = ex

y = 1 ¡ x where

Z

xex dx we integrate by parts with:

n

R ¡ x 1¢

) xex y =

Z For

³x + 1´

dy + xex y = ex (x ¡ x2 ) dx x dy ) xex + ex (x + 1)y = ex (x ¡ x2 ) dx d ) (xex y) = ex (x ¡ x2 ) dx

Z

R

dy = x ¡ x2 dx

³x + 1´

) xex 1 dx

dy + dx

I(x) = e

dy + ex y = xex + e2x dx d x ) (e y) = xex + e2x dx

=

sin x dx

cos x c + x x

(x + 1)y + x

2

) ex

) ex y =

v=x v0 = 1

¢

1 e¡2x + c ¡2 ¡ 12 e¡2x+3x + ce3x ¡ 12 ex + ce3x

But when x = 0, y = 2 ) 2=

u0 = cos x u = sin x

= x sin x ¡ (¡ cos x) + c

dy ¡ 3ye¡3x = ex e¡3x dx d ) (ye¡3x ) = e¡2x dx ) y=

x cos x dx

We integrate by parts with: e¡3x

dx

fsame as (1)g

Z

¡3 dx

1 x

= eln x =x

dy + y = x cos x dx d (xy) = x cos x dx

) e¡3x

¡

x

R

y = cos x which has I(x) = e

n

+c

e

has I(x) = e

) ye¡3x =

³1´

) xy =

¡4x

dy ¡ 3y = ex b dx

dy + dx

) x

gives

dy + 4ye4x = 12e4x dx d ) (ye4x ) = 12e4x dx ) ye

dy + y = x cos x .... (1) dx

)

e4x

4x

e2 2

) y = x ¡ 1 + 12 ex + e¡x (e ¡ 12 e2 )

EXERCISE O a

c e

e c = 2 e

) c=e¡

x h x i ) y = x ln 1 ¡ cx

1

e2x dx

ex dx +

= xex ¡ ex + 12 e2x + c

y 1 ¡c ¡ = ln x x h 1 ¡ cx i ) y = ¡x ln x ) y = x ln

Z

) ex y = xex ¡

black


v = 1 ¡ 2x v 0 = ¡2

R

(¡2ex ) dx]


WORKED SOLUTIONS ) xex y = ex (x ¡ x2 ) ¡ ex (1 ¡ 2x) ¡ 2ex + c

But y 0 (x) = 2x +

1 ¡ 2x 2 c ¡ + x x xex 1 2 c = 1¡x¡ +2¡ + x x xex 3 c = 3¡x¡ + x xex

) y = 1¡x¡

a )

) Tn (x) = ¡1 + 1(x ¡ 1) +

) Tn (x) = ¡1 + x ¡ 1 + 2x2 ¡ 4x + 2

dy e¡y = ¡1 dx 3 ³ ´ d2 y 1 ¡y dy = ¡e 3 dx2 dx dy = ¡ 13 e¡y dx

µ

=

1 ¡y e 3

b

+ e¡y

¶

d2 y dx2

¡

d2 y dx2

1 3

¡ 1 = ¡ 23

¶

)

) T3 (x) = ¡ 23 x + 19 x2 +

1 + 2e¡x 3

¶

, ey =

1 3

3 + 23 e¡x

.... (1) ) 1

) x

1

When x = 1, y(1) = 0, y 0 (1) = 3, y 00 (1) =

)

T3 (x) = y(1) + y0 (1)(x ¡ 1) +

)

d3 y = 0, etc dx3

+

=0

6(x ¡ 1)2 24(x ¡ 1)3 + 2 6

yellow

95

100

50

75

25

0

= 3(x ¡ 1) ¡ 3(x ¡ 1)2 + 4(x ¡ 1)3

5

95

y00 (1)(x ¡ 1)2 2!

y000 (1)(x ¡ 1)3 3!

= 0 + 3(x ¡ 1) ¡

100

50

75

25

0

5

95

50

75

100

magenta

3¡3(3) 1

= ¡6

d2 y =4 dx2

25

0

d3 y d2 y = ¡4 dx3 dx2

¡4(¡6) and y000 (1) = = 24 1

5

95

100

50

75

d2 y d3 y d2 y +x = ¡3 2 3 dx dx dx2 ) x

y00 (1)(x ¡ 1)2 Tn (x) = y(1) + ¡ 1) + 2! fas all other terms are zerog

25

d2 y dy = 3¡3 dx2 dx

and differentiating again

y 0 (1)(x

0

dy d2 y dy +x = 3¡2 dx dx2 dx

ffrom (1)g

dy dy +x = 4x + dx dx2 dx

cyan

X

dy 3x ¡ 2y = dx x dy ) x = 3x ¡ 2y dx

d2 y

dx(n)

¡1 = 2 + c c = ¡3

y dy = 4x ¡ 3 = 2x + (2x ¡ 3) = 2x + dx x

a

dy e¡y = ¡1 X ) dx 3 dy y = 2x + dx x dy ) x = 2x2 + y dx

d(n) y

1 x

1 dy 1 ¡ y=2 x dx x2 ³ ´ d y ) =2 dx x Z y ) = 2 dx = 2x + c x

Check:

1 3 x 81

dy = ¡ 23 e¡x dx dy ) ey = 13 ¡ ey dx dy 1 = ¡1 ) dx 3ey

) 1

= e¡ ln x

But when x = 1, y = ¡1 ) ) ) y = 2x2 ¡ 3x

2 9

) ey

a

1

¡ x dx

y = 2x2 + cx

¡ ¢ 2 d3 y = 13 49 ¡ 29 = 27 3 dx x2 x3 Now T3 (x) = f (0) + f 0 (0)x + f 00 (0) + f 000 (0) 2! 3! µ

x

R

y = 2x has I(x) = e

=

d2 y dy = ¡ 13 = dx2 dx

b If y = ln

³1´

¡1

dx

dx dy = dx

dy ¡ dx

= eln x

³ dy ´2

µ³ ´2 dy

At x = 0, y = 0,

5

4(x2 ¡ 2x + 1) 2

) Tn (x) = 2x2 ¡ 3x

d3 y = ¡ 13 ¡e¡y dx3

2

y ) y0 (1) = 2 + y x ) 1=2+y ) y = ¡1

EXERCISE P 1

199

black



200

WORKED SOLUTIONS 3x ¡ 2y 2 dy = =3¡ y dx x x

b As

R

where I(x) = e

2 x

dy 2 + y=3 dx x

then

1 dy sin x 1 + y= cos x dx cos2 x cos2 x ³ y ´ d ) = sec2 x dx cos x

)

dx

= e2 ln x 2

= eln x = x2

y = cos x

)

dy + 2xy = 3x2 dx d 2 ) (x y) = 3x2 dx

c x2

Note:

But when x = 1, y = 0 ) 0 = 1 + c ) c = ¡1 1 x2

y =x¡

sin x = x ¡

x3 x5 + ¡ :::: 3! 5!

cos x = 1 ¡

x2 x4 + ¡ :::: 2! 4!

)

sin x + 2 cos x ¼x¡

dy + y sin x = 1, y(0) = 2 dx Differentiating with respect to x gives dy d2 y dy ¡ sin x + cos x 2 + sin x + y cos x = 0 dx dx dx

a cos x

µ

) cos x

d2 y +y dx2

¶

p(p ¡ 1)x2 p(p ¡ 1)(p ¡ 2)x3 + + :::: 2! 3! dy 2p(p ¡ 1)x 3p(p ¡ 1)(p ¡ 2)x2 =0+p+ + + :::: ) dx 2! 3! p(p ¡ 1)(p ¡ 2)x2 + :::: = p + p(p ¡ 1)x + 2! dy p(p ¡ 1)(p ¡ 2)x3 ) x = px + p(p ¡ 1)x2 + + :::: dx 2! dy dy dy = +x , ) (1 + x) dx dx dx

= 0 for all x

y = 1 + px +

dy dy + 2(0) = 1 ) = 1, dx dx d2 y = ¡2 dx2

fa sum of two convergent seriesg = p + [p(p ¡ 1) + p]x

d3 y = ¡1 dx3 x2 x3 Now T3 (x) = y(0) + y0 (0)x + y 00 (0) + y 000 (0) 2! 3!

·

and

+

·

dy + y sin x = 1 dx ³ sin x ´ dy 1 + y= dx cos x cos x

cos x

R

where I(x) = e

R

¡

=e

sin x cos x

=

¡ sin x cos x

yellow

95

50

75

25

0

5

95

100

50

75

25

0

5

95

100

magenta

¸

1 p(p ¡ 1) :::: (p ¡ n + 1)xn P dy = p+p dx n! n=1

= py

75

50

1 cos x

25

0

5

95

100

50

75

25

0

5

p(p ¡ 1) :::: (p ¡ n + 1) n x n!

(1 + x)

=e

cyan

·

Hence

dx

¡ ln(cos x)

=

p(p ¡ 1) :::: (p ¡ n + 1) [(p ¡ n) + n]xn n!

=p

dx

¸

p(p ¡ 1) :::: (p ¡ n) p(p ¡ 1) :::: (p ¡ n + 1) n + x n! (n ¡ 1)!

as required

100

)

¸

p(p ¡ 1)(p ¡ 2) + p(p ¡ 1) x2 + :::: 2!

where the general term is:

) T3 (x) ¼ 2 + x ¡ x2 ¡ 16 x3

b

+ ::::

a For ¡1 < x < 1,

5

But when x = 0, y = 2 ) cos 0

1 4 x 12

which checks with the answer in a.

d3 y dy =¡ dx3 dx

and

x3 x5 2x2 2x4 + +2¡ + + :::: 3! 5! 2! 4!

¼ 2 + x ¡ x2 ¡ 16 x3 +

d2 y = ¡y dx2

)

cos x

´

+c

But when x = 0, y = 2 ) 2 = 0 + c ) c=2 ) y = sin x + 2 cos x

3x2 dx = x3 + c

) y =x+

4

³ sin x

) y = sin x + c cos x

Z

)

sec2 x dx = tan x + c

) y = cos x

) x2

) x2 y =

Z

black



WORKED SOLUTIONS b

dy = py dx 1 dy p = y dx 1+x

(1 + x)

Z

1 dy = y

)

Z

n!1

= lim

n!1

= =

for some constant A.

d

p(p ¡ 1) :::: (p ¡ n + 1)xn = (1 + x)p n!

n=0

But )

REVIEW SET A 1 ln x and x are continuous for all x > 0 The limit has type

5


ln x = lim x x!1 x x

lim

x!1

1 P 1 n=1 1

P


2 As x ! 0, ex sin x ! 0 also The limit has type )

lim

5 6+ n ¡ n72


n2

3x + x¡2

x!0

j3xj < jx ¡ 2j j3xj2 < jx ¡ 2j2 9x2 ¡ (x ¡ 2)2 < 0 (3x + x ¡ 2)(3x ¡ x + 2) < 0 (4x ¡ 2)(2x + 2) < 0

¡ +

=

0¡0¡2 0+0+7

=

¡ 27

2 n 6 n

cyan

b

´ 1 ³ P 3x n

n=0

+7

Qw

x

1 2

=

u1 1¡r 1 1¡

3x x¡2

£

x¡2 x¡2

x¡2 x ¡ 2 ¡ 3x x¡2 = ¡2x ¡ 2 2¡x = f¡1 < x < 12 g 2x + 2 =

1 + 2n which diverges as n ! 1 n

magenta

yellow

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

x¡2

=

1 + n[1 + (¡1)n ] diverges n

5

95

100

50

75

25

0

3+

+

Note: The ratio test could have also been used.

¡2

b If n is even, un = 3 +

5

-1

4 + 6n + 7n2

8 n2 = lim n!1 4 n2

´

+

)

) ¡1 < x < lim

´2 ³

+

8 ¡ 2n ¡ 2n2

n!1

)

3x x¡2


=1 a

³

fComparison testg

)

= 1(0) + 1(1)

3

fp-series testg

converges

n converges n3 + 1

) ) )

ex sin x

x ex sin x + ex cos x = lim x!0 1

¼ 2

3 3x + :::: is a geometric x¡2 3x series with u1 = 1 and r = x¡2 ¯ ¯ ¯ 3x ¯ < 1 ) the series converges when jrj = ¯ ¯ x¡2

=0

0 , 0

q

2 + 13 n

lim arctan n =

a 1+

1

)

5 6+ n ¡ n72

n

2 p 6 p 6 3

n=1

1 , 1

2n + 13

1 2 3 4 + + + + :::: 13 + 1 23 + 1 33 + 1 43 + 1 1 1 1 P P P n n 1 = which is 6 = 3 3 2 n + 1 n n n=1 n=1 n=1

4

for all jxj < 1, that is x 2 ]¡1, 1[ .

´

q

n!1

Now y(0) = 1 ) 1 = A(1 + 0)p ) A=1 1 P

³

5 n2 6 + n ¡ n72

= lim

p(p ¡ 1) :::: (p ¡ n + 1)xn = A(1 + x)p n!

Thus y(x) =

r

2n + 13

6n2 + 5n ¡ 7 2n + 13

p dx 1+x

solution to the differential equation,

n=0

= lim

n!1

Since jxj < 1, we have y = A(1 + x)p , where A = ec is a constant. 1 P p(p ¡ 1) :::: (p ¡ n + 1)xn is a c Since y(x) = n! n=0 1 P

p

n!1

) ln jyj = p ln j1 + xj + c, where c is a constant

y(x) =

lim

c

201

black



202

WORKED SOLUTIONS c Differentiating the result in b with respect to x:

´ 1 ³ d P 3x n = dx n=0 x ¡ 2 µ

1 P

)

³

d 3x dx x ¡ 2

n=0

ń ¶

n

x¡2

n=0

³

´

1 2 1 2

d 2¡x dx x + 1

· =

³ 3x ń¡1 1 P n

x¡2

n=0

)

6 x+

´

1 2

=

n+1 n=0 (x ¡ 2)

·

1 2

1 P

an converges for

) ) ) )

¡3 (x + 1)2

dy x = = k where k = 0, §1, §4 dx y ) we have isoclines x = 0, y = §x, y = § 14 x 10

x , but we will 1¡x

)

Z

2

sin(x ) dx = 0

¼

1 dy dx = y dx

)

1 3

1 42

+

1 1320

¯ ¯

) ln ¯¯

x=0

x!

= e¡¸ =e

¸1

11

dy ¡ dx

x=0

x!

³1´ x

R

p x has I(x) = e

1

¡ x dx

= e¡ ln x ¡1

= eln x =

1 x

1 1 dy 1 ¡ ¡ 2 y=x 2 x dx x ³ ´ 1 d y ¡ ) =x 2 dx x Z 1 y ¡ ) = x 2 dx x

)

1

is the Maclaurin series for e¸ g )

y x2 = 1 +c x 2

magenta

yellow

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

p ) y = 2x x + cx

75

25

0

5

95

100

50

75

25

0

5

y=

y = 2(x ¡ 1)ex¡2

1 75 600

= e0 =1

cyan

)

0

x!

1 P ¸x

y = §ex¡2 2(x ¡ 1)

But x = 2, y = 2 does not satisfy the negative solution

e

fas

¯

¯ y ¯=x¡2 2(x ¡ 1) ¯

)

1 P ¸x

x=0 ¡¸ ¸

ln 2 = 2 + c c = ln 2 ¡ 2

¯ ¯ ¯ y ¯ x¡2 ¯ ¯ ¯ 2(x ¡ 1) ¯ = e

)

e¡¸ ¸x 8 P(X = x) = where x = 0, 1, 2, 3, 4, .... x! 1 P e¡¸ ¸x

´

1 dx x¡1

) ln jyj ¡ ln j2(x ¡ 1)j = x ¡ 2

1 2

¡

1+

) ln jyj = x + ln jx ¡ 1j + ln 2 ¡ 2

x3 x7 x11 x15 ¡ + ¡ + :::: 3 42 1320 75 600 ¡

Z ³

) ln jyj = x + ln jx ¡ 1j + c

¼ 0:310 Check: GDC gives ¼ 0:310 268

)

dy xy = dx x¡1 x dy = dx x¡1 x¡1+1 1 dy = =1+ dx x¡1 x¡1

But, when x = 2, y = 2 ) )

x6 x10 x14 x18 + ¡ + ¡ :::: 3! 5! 7! 9!

·

1

)

1 y 1 y

)

x3 x5 x7 x9 + ¡ + ¡ :::: 3! 5! 7! 9!

) sin(x2 ) = x2 ¡

y = -x

x=0

jxj2 < j1 ¡ xj2 2 x ¡ (1 ¡ x)2 < 0 (x + 1 ¡ x)(x ¡ 1 + x) < 0 2x ¡ 1 < 0

sin(x) = x ¡

y = -Vr

¸

x 1¡x

x
1 + r2

r=1 1

r

2

1 2k

´ i ¼

=1

2k

diverges.

P 1+r 1 + r2

a Sn =

1 X 1

r r=1

diverges

n P k=3

diverges

fp-series testg fComparison testg

(¡1)k+1 ln(k ¡ 1)

Consider bk =

1 where k > 3 ln(k ¡ 1)

yellow

95

50

75

25

0

5

95

100

50

75

25

0

Now y = ln x is an increasing function for all x > 0.

5

95

100

50

75

25

0

5

95

³

¡2

¡¼¢

¡

1+r 1 > 1 + r2 r

P1

r=1

¡2 = ¡ 19 lim un = n!1 3£6

100

50

75

25

0

5

)

1 n

1 2

r + r2 > 1 for all r > 1 1 + r2

4

100

p

h³

¶

fTest for divergenceg

(n + 5) ¡ (n ¡ 1) p n+5+ n¡1

where

(k ¡ 1)¼ 2k

³ (k ¡ 1)¼ ´

= p

)

1 n and bn ! 0

alternating series with bn =

a If n is even, un =

c un =

provided jxj < 1

an

¡1 < x < 1 and is divergent for jxj > 1.

-1

1

n+1

n!1

= jxj

4

-2

203

black



204

WORKED SOLUTIONS ) )

1 is decreasing for all x > 0 ln x bk is decreasing for all k > 3

The equation of the tangent is y = mx + c dy ) y= x + 3x2 y3 dx dy = y ¡ 3x2 y3 fsee Exercise N question 8g ) x dx dy dv Let y = vx, so = x+v dx dx

And, as bk > 0 for all k > 3 n P

Sn =

k=3

(¡1)k+1 converges fAlternating series testg ln(k ¡ 1)

dv + vx = vx ¡ 3x2 v 3 x3 dx dv ) x2 = ¡3x2 v 3 x3 dx 1 dv ) = ¡3x3 v3 dx Z Z dv ) v ¡3 dx = ¡3 x3 dx dx

b Now jS ¡ S10 j 6 b11 when S = lim Sn

Hence, x2

n!1

fAlternating Series Estimation Theoremg ) jS ¡ S10 j 6

1 ln 10

) jS ¡ S10 j 6 0:4343 f4 s.f.g 6 The Taylor series expansion of ex is: ex = 1 + x +

x2 x3 x4 + + + :::: 2! 3! 4!

) ex¡1 = 1 + (x ¡ 1) +

(x ¡ 1)4

+

4!

(x ¡ 1)2 2!

+

µ ¶2

3!

)

+ ::::

)

) y¡2

)

Z

dy dx = dx

)

=

c = 10

3 4 x 2

¡ 20

x2 3x4 ¡ 40 = y2 2

)

) y2 =

2x2 ¡ 40

3x4

r

) y=§

Z

2x2 (3x4 ¡ 40)

p8

When x = 2, y = §

r

(2x ¡ 1) dx

y=

Z

y ¡2 dy =

)

x y

)

dy = 2xy2 ¡ y 2 = (2x ¡ 1)y2 dx 1 dy = 2x ¡ 1 y 2 dx

Z

= ¡ 34 x4 + c

¡ 12 (4) = ¡12 + c

µ ¶2

dy = 4x ¡ 2y dx

then a = 4x ¡ 2(ax + b) for all x ) a = (4 ¡ 2a)x ¡ 2b for all x ) 4 ¡ 2a = 0 and a = ¡2b ) a = 2 and b = ¡1 8

x y

¡ 12

But when x = 2, y = 1

) (x ¡ 1)ex¡1 ¼ (x ¡ 1) + (x ¡ 1)2 + 12 (x ¡ 1)3 7 If y = ax + b is a solution of

v¡2 ¡3x4 = +c ¡2 4

)

(x ¡ 1)3

8

= §1 indicates that the solution is

2x2 . 3x4 ¡ 40

(2x ¡ 1) dx

y ¡1 2x2 = ¡x+c ¡1 2 ¡1 ) = x2 ¡ x + c y ¡1 ) y= 2 x ¡x+c

10 From equation a,

)

)

dy = 1 at (0, 0) dx

a has slope field B.

From equation b, )

dy = 0 at (2, 2) dx

b has slope field C.

Consequently c has slope field A.

9

y

11

(x, y)

a

1 1 400 ¡ P + P + = P 400 ¡ P P (400 ¡ P ) =

3x2y3

400 P (400 ¡ P )

(2, 1)

b Notice that as P is growing

³

x

cyan

magenta

yellow

95

400

dP >0 dt

>0

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25

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5

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50

75

25

0

5

95

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50

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25

0

5

95

100

50

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25

0

5

) 0:2P 1 ¡

´ P

black



WORKED SOLUTIONS )

P (400 ¡ P ) >0 400

12

) P (400 ¡ P ) > 0

) f (0) =

Sign diagram of P (400 ¡ P ) is: -

+

=

-

0

)

µ ³

´

b

dP P = 0:2P 1 ¡ dt 400 ³ 400 ¡ P ´ dP ) = 15 P dt 400 400 dP = 15 P (400 ¡ P ) dt Now as

) ) )

Z ³ 1

P

+

1 400 ¡ P

1 + P 400 ¡ P 1 ¡1

) ln jP j +

´ dP

´ dP

=

dt

1 5

ln j400 ¡ P j =

µ

=

³

P 400 ¡ P

´

=

)

t 400 ¡ P ¡ ¡c =e 5 P

1 5

=

1 t 5

dt

+c

=

¡

1 + Ae

)

c When t = 20, P = t ¡5

d As t ! 1, e )

1 P

where r =

ri

t 5

x2 1 + x2

1 P

¯ ¯ ¯1 + x2 ¯

¯

¯

)

1 < ¯1 + x2 ¯

) ) )

1 < 1 + x2 0 < x2 x 6= 0

d For x = 0, f (0) = 0 from a. For x = 6 0, 1 x2 P ri 1 + x2 i=0

x2 1 + x2

x2 = 1 + x2

¼ 389 people

=

yellow

³ 1 ´

fGeometric seriesg

1¡r

0

1

x2 1 + x2

95

f (x) =

µ µ

1 + x2 1 + x2 ¡ 1 1 + x2 x2

n 1, 0,

¶

¶

x 6= 0 x=0

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25

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5

95

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50

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25

0

5

95

0 and 400 ¡ P > 0 ffrom ¤g ) ln

x2 (1 + x2 )i i=1

1 + x2 x2

1 + x2 = x2

Z

dt =

dt

1 P

i=1

P

400

0 ln n for all n 2 Z +

1 1 < for all n 2 Z + n ln n 1 1 P P 1 1 ) < n ln n n=2 n=2 )

REVIEW SET C 1

p

n2

a n¡

+n=

=

³

p

n2

n¡

³

n¡

n!1

+n

n+ n+

p

n2 + n

!

p

n2 + n

¡n

´

p

)

n2 + n

q

h

n2 + n

1 n

1+

Thus 4

¡n

q

h

1 1+ n

n 1+

n!1

1+

q

1 1+ n

fsince n 6= 0g =

¡1 1+1

P 1

fsince

lim

n!1

1 = 0g n

= ¡ 12

P

1 diverges 2) ln(n n=2

) both

b

n

n

n

1

(3n ) n

)

6 (3 + n

n

1

1

2n ) n

3n ) n

) 31 6 (3n + 2n )

1 n

6 (2 £

1 P

6 3£2

1 P

)

1 ³ P

an ¡

1

converges .... (¤)

n=1

1 n

´2

lim

Thus

n!1

= 3 fSqueeze theoremg

an = 0 for all a > 0 fTheoremg, n!1 n! en lim =0 n!1 n!

d Let un =

(¡1)n ne¡n

If n is even, If n is odd,

=

(¡1)n

(2) is convergent as

3

1 r

Thus

an ¡

1 P

1 r

5

magenta

a

n2

| {z } n=1

(3)

1 P

an 6 n

1 P

an

fComparison testg

n=1

1 n

´2

converges.

an2 is divergent.

95

1 x(x + 1)

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25

0

5

95

yellow

1 P

(¡1) p . n

1 1 x+1¡x ¡ = x x+1 x(x + 1) =

100

50

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25

0

5

95

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50

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25

0

5

95

(2)

n=1

1 1 P

n=1

1

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50

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25

0

(1)

n=1

r=1

5

| {z }

an is convergent, but

3 r is not convergent fTest of divergenceg

cyan

n

+

| {z }

For a counter example, consider an =

lim ar = 30 = 1 6= 0

1 P

1 a P n

b No, we can only apply the Limit Comparison test if an > 0 for all n. n

r!1

)

1 ³ P

n=1

r=1

Now

an2 ¡ 2

(3) is convergent fp-series testg

n un = n ! 0+ as n ! 1 e ¡n un = n ! 0¡ as n ! 1 e

we let ar = 3

P

n=1

n en

n!1

1 P

n=1 1

´

(1) is convergent ffrom ¤g

lim (¡1)n ne¡n = 0

)

2an 1 + 2 n n

an2 ¡

n=1

lim

c As

2 For

+

1 ³ P

= =

1

2n ) n

bn converge

n=1

fLimit Comparison testg an2

n=1

1 n

where 2 n ! 1 as n ! 1 (3n

ffrom ¤ g

1 P

an and

n=1

Now 3 6 3 + 2 6 3 + 3 n

fComparison testg

diverges

ln n

a We use the Limit Comparison test which states: “If an > 0 and bn > 0 for all n 2 Z + and if bn lim = c where c is a real constant, then an and bn n!1 an either converge or diverge together.” Suppose lim an = L and consider bn = an2 n!1 bn As an > 0 and an2 > 0 and = an an bn then lim = lim an = L n!1 an n!1

i

¡1

= lim

fp-series testg

diverges

n

n=2 1 n=2 1

i

= lim

n!1

1 P 1

But

p

n+

n 1+ lim

Ã

n2 ¡ (n2 + n)

=

)

´

black



WORKED SOLUTIONS b Consider f (x) =

1 . x(x + 1)

Thus f (n+1) (c) =

If x > 1, f (x) is > 0 and is continuous and

Z

1

1

1 dx = lim b!1 x(x + 1)

Z b³ 1 1

£

= lim = lim = lim

b!1

dx

ln

x+1

³ ³ b ´ ln

b+1

Ã Ã

1

= lim

ln

b!1

1+

1

¡ ln

where

2

Thus jRn (x : 0)j =

!

=

+ ln 2

1 b

Hence,

n=1

c

i

= ln 2 1 is convergent fIntegral testg n(n + 1)

Sn

P n

=

i=1

=

i=1

³

¡

i

1 1

1 i+1

ffrom ag

´

+

³

´

³

´

Thus

Z ) )

1 P

1 ii = lim Sn n(n + 1) n!1 n=1

1 P

)

lim

n!1

1 =0 n+1

x0 x1 x2 x3 x4 x5 x6

yellow

95

50

75

25

0

5

95

100

50

75

25

0

)

5

95

100

50

75

(¡1)n n! (1 + x)n+1

25

0

5

95

100

50

75

0

¸ n+1 x

t n+1

0

= ln(1 + x) ¡ ln 1

0

1 P (¡1)n xn+1

n+1 x2 2

+

x3 x4 x5 ¡ + ¡ :::: 3 4 5

Now xn+1 = xn + 0:1 and yn+1 = yn + 0:1f (xn , yn ) = yn + 0:1(xn ¡ 2yn ) = 0:1xn + 0:8yn

) f (n+1) (x) = (¡1)(¡2)(¡3) :::: (¡n)(1 + x)¡n¡1

25

¤x

dy = x ¡ 2y and when x = 1, y = 2 dx

8

f (4) (x) = (¡1)(¡2)(¡3)(1 + x)¡4

0

0

1 dt 1+t

tn dt = ln(1 + t)

n+1

f 000 (x) = (¡1)(¡2)(1 + x)¡3

5

£

x

=x¡

1 = (1 + x)¡1 1+x

magenta

(¡1)n

(¡1)n

x

n=0

f 00 (x) = ¡1(1 + x)¡2

cyan

Z

(¡t) dt =

Z

·

1 for ¡1 < x < 1 1+x

n

n=0

n=0

f (x) = ln(1 + x), 0 6 x < 1

) f (n+1) (x) =

P

´ 1

1¡

= 1, since

(¡x)n =

) ln(1 + x) =

³

n!1

) f 0 (x) =

1 n+1

=1 = 1:1 = 1:2 = 1:3 = 1:4 = 1:5 = 1:6

y0 y1 y2 y3 y4 y5 y6

=2 = 1:7 = 1:47 = 1:296 = 1:1668 = 1:073 44 = 1:008 752

y6 ¼ 1:009 fto 4 s.f.g

100

= lim

0

1 P

n=0 x 1

n=0

1 P

6

1+c

u1 for jrj < 1 1¡r 1 = for jxj < 1 1+x

So, its sum is

1 1 1 + 12 ¡ + 13 ¡ 1+1 2+1 3+1 ³ 1 ´ ³1 ´ 1 1 + :::: + ¡ + ¡ n¡1 n¡1+1 n n+1 1 1 1 1 1 1 1 1 = 1 ¡ 2 + 2 ¡ 3 + 3 ¡ 4 + :::: + ¡ + n¡1 n n 1 ¡ n+1 1 =1¡ n+1 =

³ x ń+1

7 1 ¡ x + x2 ¡ x3 + x4 ¡ x5 + :::: is a geometric series with u1 = 1 and r = ¡x

1 i(i + 1)

n P 1

n! jxjn+1 n+1 (1 + c) (n + 1)!

But 0 6 x < 1 and 1 < 1 + c < 2 x ) 06 61 1+c ³ ń+1 x 1 fas ! 0g ) jRn j 6 n+1 1+c

= ln 1 + ln 2 1 P

¯ ¯ ¯ f (n+1) (c)xn+1 ¯ ¯ ¯ jRn (x : 0)j = ¯ ¯ (n + 1)!

and c lies between 0 and 1 ) 1 < 1 + c < 2

¡ 1 ¢´

!

1

first n terms of Taylor series

¤b

h ³ x íb

(¡1)n n! (1 + c)n+1

Now ln(1 + x) = Tn (x) + Rn (x : 0)

´

ln x ¡ ln(x + 1)

b!1

b!1

1 ¡ x x+1

207

black



208

WORKED SOLUTIONS a If V is the volume of water remaining then 4 ¡ V escaped.

9

Z

has

)

dv dx = dx

v

v2 = ln jxj + c 2

)

y2 = 2 ln jxj + 2c x2

)

V = volume remaining

11

p d (4 ¡ V ) / h dt 1 dV ) ¡ = kh 2 dt

) sin x

4 +c k 1

But when t = 0, h = 1 ) 0 = ¡ ) c=

¼ , 2

8 k

a un =

1

8 (0:1) k

Consider an = ) an+1 ¡ an =

p h) fin (2)g

sin x

2n + 1 3n ¡ 1

2(n + 1) + 1 2n + 1 ¡ 3(n + 1) ¡ 1 3n ¡ 1

=

6n2 + 7n ¡ 3 ¡ 6n2 ¡ 7n ¡ 2 (3n + 2)(3n ¡ 1)

=

¡5 (3n + 2)(3n ¡ 1)

magenta

yellow

fan g is strictly decreasing 7 9 11 , , , 8 11 14

95

) )

.... decrease

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

) an+1 ¡ an < 0 for all n ) an+1 < an for all n

75

50

¡ 14 (2 cos2 x)

=

dv dy = x+v dx dx

25

0

5

95

100

50

cyan

= ¡ 14

2n + 3 2n + 1 ¡ 3n + 2 3n ¡ 1 (2n + 3)(3n ¡ 1) ¡ (2n + 1)(3n + 2) = (3n + 2)(3n ¡ 1)

when h = 0, t = 20 min

dv x vx x+v = + dx vx x dv 1 x+v = +v dx v dv 1 ) v = dx x

75

25

)

1 (¡1) 4

3 £ 5 £ 7 £ :::: £ (2n + 1) 2 £ 5 £ 8 £ :::: £ (3n ¡ 1)

x y dy = + dx y x

)

cos 2x + c

REVIEW SET D

8 ) = 20 k

Let y = vx, so

(¡ cos 2x) + c

cos2 x ) y=¡ 2 sin x

8 +c k

and when t = 2, h = 0:81 ) 2 =

10

2

y=0

) y=

p 8 Thus t = (1 ¡ h) .... (2) fin (1)g k

0

) y sin x =

¡ 14

2

) y sin x = ¡ 14 (cos 2x + 1)

2

5

) y sin x =

) c=

8p h + c .... (1) ) t=¡ k

)

sin 2x dx

) 0 = ¡ 14 cos ¼ + c

1 h2

t = 20(1 ¡

1 2

¡ ¢ 1 1

But when x =

dt 4 ¡1 =¡ h 2 dh k

)

dx

Z

) y sin x =

1 dV dh =4 = ¡kh 2 dt dt dh k 1 ) = ¡ h2 dt 4

) t=¡

cos x sin x

dy + cos x £ y = sin x cos x dx d ) (y sin x) = 12 sin 2x dx

) V = 4h

c From b,

cot x dx

=e = sin x

But V = 2 £ 2 £ h m3

)

R

dy + (cot x)y = cos x has I(x) = e R dx =e

ln(sin x)

1 dV = ¡kh 2 m3 /min dt

) b

) y 2 = 2x2 (ln jxj + c)

2m 2m

1 dx x

) hm 1m

Z

black



WORKED SOLUTIONS

¡ 3 ¢¡ 7 ¢n

Hence, un
3 ) n + 1 > 4

(0:3)3 = 0:0045 so we use n = 4 But 3!

¼ 1:350

)

c Suppose e is rational, that is, e =

dy = 1 + x + y2 where y(1) = 0 dx dy dv Let y = vx ) = x+v dx dx

positive integers.

6 xy

x+v

´

v2 2

)

8

2

) y = ¡1 ¡ 2x + 2cx But when x = 1, y = 0 ) 0 = ¡1 ¡ 2 + 2c ) 2c = 3

dy + dx

cyan

f (n+1) (c)xn+1 (n + 1)!

=

magenta

3 x

dx

3 ln x

=e = x3

8x8 +c 8 c ) y = x5 + 3 x

But when x = 1, y = 0 ) 0 = 1 + c ) c = ¡1 Thus, y = x5 ¡

yellow

95

50

75

25

0

5

95

100

50

75

8x7

) x3 y =

ec xn+1

25

0

5

95

100

50

75

0

95

100

50

75

25

0

5

k=0

x

R y = 8x4 , y(1) = 0 has I(x) = e

) x3 y =

ec 1 + , 0

Mathematics for the international student Mathematics HL (option): Calculus

Mathematics for the International Student: Mathematics HL - International

Mathematics HL for the international student (int'l baccalaureate)(2004)